Book I of the Elements ΣΤΟΙΧΕΙΩΝ Α Oumlğelerin Birinci Kitabı

Euclid ΕΥΚΛΕΙΔΟΣ Oumlklid

September

This work is licensed under theCreative Commons

Attribution-NonCommercial-ShareAlike Unported License

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or send a letter toCreative Commons

Castro Street Suite Mountain View California USA

Bu ccedilalışmaCreative Commons Attribution-Gayriticari-ShareAlike

Unported Lisansı ile lisanslıLisansın bir kopyasını goumlrebilmek iccedilin

httpcreativecommonsorglicensesby-nc-sa30

adresini ziyaret edin ya da mektup atınCreative Commons

Castro Street Suite Mountain View California USA

CCcopy BYcopy Oumlzer Oumlztuumlrk amp David Pierce $copy

C

copy

Matematik BoumlluumlmuumlMimar Sinan Guumlzel Sanatlar Uumlniversitesi

Bomonti Şişli İstanbul

ozerozturkmsgsuedutr dpiercemsgsuedutr

httpmatmsgsuedutr

This edition of the first book of Euclidrsquos Elements was prepared for a first-year undergraduate course in themathematics department of Mimar Sinan Fine Arts University The text has been corrected after its use in the coursein the fall of

Oumlklidrsquoin Oumlğelerrsquoinin bu baskısı Mimar Sinan Guumlzel Sanatlar Uumlniversitesi Matematik Boumlluumlmnde bir birinci sınıflisans dersi iccedilin hazırlanmıştır ndash Guumlz doumlneminde bu notlar ilk defa kullanılmış ve fark edilen hatalarduumlzeltilmiştir

Introduction

Layout

Book I of Euclidrsquos Elements is presented here in threeparallel columns the original Greek text in the middlecolumn an English translation to its left and a Turkishtranslation to its right

Euclidrsquos Elements consist of books each dividedinto propositions Some books also have definitions

and Book I has also postulates and common notions

In the presentation here the Greek text of each sentenceof each proposition is broken into units so that

each unit will fit on one line the unit as such has a role in the sentence the units kept in the same order make sense when

translated into EnglishEach proposition of the Elements is accompanied by

a picture of points and lines with most points (and somelines) labelled with letters This picture is the lettered

diagram We place the diagram for each proposition af-ter the words According to Reviel Netz [ p n ]this is where the diagram appeared in the original scrollpresumably so that one would know how far to unroll thescroll in order to read the proposition The end of a propo-sition is not to be considered as an undignified positionIndeed Netz judges the diagram to be a metonym forthe proposition something associated with the proposi-tion that is used to stand for the proposition (Today theenunciation of a propositionmdashsee sect belowmdashwould appearto be the common metonym)

Text

We receive Euclidrsquos text through various filters TheElements are supposed to have been composed around bce Heibergrsquos text (published in ) is based mainlyon a manuscript in the Vatican written the tenth centuryce closer to our time than to Euclidrsquos time Knorr []argues that Euclidrsquos original intent may be better reflectedin some Arabic translations from the eighth and ninth cen-turies (The argument is summarized in []) Nonethelesswe shall just use the Heiberg text

More precisely for convenience we take the Greek textin our underlying LATEX file from the LATEX files of RichardFitzpatrick who has published his own parallel Englishtranslation (In the underlying LATEX file the enunci-ation of Proposition I in Greek reads as in Table )Fitzpatrick reports that his Greek text is that of Heiberg

but he gives it without Heibergrsquos apparatus criticus Alsohis method of transcription is unclear There is at leastone mistake in his text (τρὸς for πρὸς near the beginningof I) We shall correct such mistakes if we find themalthough we shall not look for them systematically

In the process of translating we have made use of aprintout of the Greek text of Myungsunn Ryu We donot have a LATEX file for this text only pdf The text issaid to be taken from the Perseus Digital Library

We also refer to images of Heibergrsquos original text []which are available as pdf files from the Wilbour Hallwebsite and from European Cultural Heritage Online(ECHO) In preparing the files from the latter source forprinting we have trimmed the black borders by means ofa program called briss

gtEplsquoi t~hc dojersquoishc egtujersquoiac peperasmrsquoenhc trrsquoigwnon gtisrsquoopleuron sustrsquohsasjai

Table Greek text coded for LATEX

Analysis

Each proposition of the Elements can be understoodas being a problem or a theorem Writing around ce Pappus of Alexandria [ pp ndash] describes the

distinction

Those who favor a more technical terminology ingeometrical research use

httpfarsidephutexasedueuclidhtmlhttpenwikipediaorgwikiFileEuclid-Elementspdfhttpwwwwilbourhallorghttpechompiwg-berlinmpgdehome

httpbrisssourceforgenetIvor Thomas [ p ] uses inquiry here in his translation

but there is no word in the Greek original corresponding to this orto proposition

bull problem (πρόβλημα) to mean a [proposi-tion] in which it is proposed to do or con-struct [something] and

bull theorem (θεώρημα) a [proposition] inwhich the consequences and necessary im-plications of certain hypotheses are investi-gated

but among the ancients some described them allas problems some as theorems

In short a problem proposes something to do a the-orem proposes something to see (The Greek for theoremmeans more generally lsquothat which is looked atrsquo and is re-lated to the verb θεάομαι lsquolook atrsquo from this also comesθέατρον lsquotheaterrsquo)

Be it a problem or a theorem a propositionmdashor moreprecisely the text of a propositionmdashcan be analyzed intoas many as six parts The Green Lion edition [ p xxiii]of Heathrsquos translation of Euclid describes this analysis asfound in Proclusrsquos Commentary on the First Book of Eu-clidrsquos Elements [ p ] In the fifth century ceProclus writes

Every problem and every theorem that is fur-nished with all its parts should contain the fol-lowing elements

) an enunciation (πρότασις)) an exposition (ἔκθεσις)) a specification (διορισμός)) a construction (κατασκευή)) a proof (ἀπόδειξις) and) a conclusion (συμπέρασμα)

Of these the enunciation states what is given andwhat is being sought from it for a perfect enunci-ation consists of both these parts The expositiontakes separately what is given and prepares it inadvance for use in the investigation The specifica-tion takes separately the thing that is sought andmakes clear precisely what it is The construc-tion adds what is lacking in the given for findingwhat is sought The proof draws the proposed in-ference by reasoning scientifically from the propo-sitions that have been admitted The conclusionreverts to the enunciation confirming what hasbeen proved

So many are the parts of a problem or a theoremThe most essential ones and those which are al-ways present are enunciation proof and conclu-sion

Alternative translations are

bull for ἔκθεσις setting out andbull for διορισμός definition of goal [ p ]Heibergrsquos analysis of the text of the Elements into

paragraphs does not correspond exactly to the analysisof Proclus but Netz uses the analysis of Proclus in hisShaping of Deduction in Greek Mathematics [] and weshall use it also according to the following understanding

The enunciation of a proposition is a general state-ment without reference to the lettered diagram Thestatement is about some subject perhaps a straight lineor a triangle

In the exposition that subject is identified in thediagram by means of letters the existence of the subjectis established by means of a third-person imperative verb

(a) The specification of a problem says what will bedone with the subject and it begins with the words δεῖ δὴHere δεῖ is an impersonal verb with the meaning of lsquoit isnecessary torsquo or lsquoit is required torsquo or simply lsquoone mustrsquowhile δή is a lsquotemporal particlersquo with the root meaning oflsquoat this or that pointrsquo [] That which is necessary is ex-pressed by a clause with an infinitive verb In translatingwe may use the English form lsquoIt is necessary for A to beBrsquo(b) The specification of a theorem says what will beproved about the subject and it begins with the wordsλέγω ὅτι lsquoI say thatrsquo The same expression may also ap-pear in a problem in an additional specification at thehead of the proof after the construction

In the construction if it is present the second wordis often γάρ a lsquoconfirmatory adverb and causal conjunc-tionrsquo [ para p ] We translate it as lsquoforrsquo at the be-ginning of the sentence but again γάρ itself is the secondword because it is postpositive it simply never appearsat the beginning of a sentence

Then the proof often begins with the particle ἐπείlsquobecause sincersquo The ἐπεί (or other words) may be fol-lowed by οὖν a lsquoconfirmatory or inferentialrsquo postpositiveparticle [ para p ]

The conclusion repeats the enunciation usuallywith the addition of the postpositive particle ἄρα lsquothere-forersquo Then after the repeated enunciation the conclusionends with one of the clauses(a) ὅπερ ἔδει ποιῆσαι lsquojust what it was necessary to dorsquo(in problems) Heiberg translates this into Latin as quodoportebat fieri although quod erat faciendum or qef isalso used(b) ὅπερ ἔδει δεῖξαι lsquojust what it was necessary to showrsquo (intheorems) in Latin quod erat demonstrandum or qed

Language

The Greek language that we have begun discussing isthe language of Euclid ancient Greek This language be-longs to the so-called Indo-European family of languagesEnglish also belongs to this family but Turkish does not

However in some ways Turkish is closer to Greek thanEnglish is Modern scientific terminology in English orTurkish often has its origins in Greek

Writing

Proclus was born in Byzantium (that is Constantinople nowİstanbul) but his parents were from Lycia (Likya) and he was ed-

ucated first in Xanthus He moved to Alexandria then Athens tostudy philosophy [ p xxxix]

capital minuscule transliteration nameΑ α a alphaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z zetaΗ η ecirc etaΘ θ th thetaΙ ι i iotaΚ κ k kappaΛ λ l lambdaΜ μ m muΝ ν n nuΞ ξ x xiΟ ο o omicronΠ π p piΡ ρ r rhoΣ σv ς s sigmaΤ τ t tauΥ υ y u upsilonΦ φ ph phiΧ χ ch chiΨ ψ ps psiΩ ω ocirc omega

Table The Greek alphabet

The Greek alphabet in Table is the source for theLatin alphabet (which is used by English and Turkish)and it is a source for much scientific symbolism The vow-els of the Greek alphabet are α ε η ι ο υ and ω whereη is a long ε and ω is a long ο the other vowels (α ιυ) can be long or short Some vowels may be given tonalaccents (ά ᾶ ὰ) An initial vowel takes either a rough-breathing mark (as in ἁ) or a smooth-breathing mark (ἀ)the former mark is transliterated by a preceding h andthe latter can be ignored as in ὑπερβολή hyperbolecirc hy-perbola ὀρθογώνιον orthogocircnion rectangle Likewise ῥ istransliterated as rh as in ῥόμβος rhombos rhombus Along vowel may have an iota subscript (ᾳ ῃ ῳ) especially

in case-endings of nouns Of the two forms of minusculesigma the ς appears at the ends of words elsewhere σvappears as in βάσις basis base

In increasing strength the Greek punctuation marksare corresponding to our (The Greekquestion-mark is like our semicolon but it does not ap-pear in Euclid)

Euclid himself will have used only the capital lettersthe minuscules were developed around the ninth century[ para p ] The accent marks were supposedly inventedaround bce because the pronunciation of the ac-cents was dying out [ para p ]

Nouns

As in Turkish so in Greek a single noun or verb canappear in many different forms The general analysis isthe same the noun or verb can be analyzed as stem +ending (goumlvde + ek)

Like a Turkish noun a Greek noun changes to showdistinctions of case and number Unlike a Turkish nouna Greek noun does not take a separate ending (such as-ler) for the plural number rather each case-ending hasa singular form and a plural form (There is also a dualform but this is rarely seen although the distinction be-tween the dual and the plural number occurs for examplein ἑκάτεροςἕκαστος lsquoeithereachrsquo)

Unlike a Turkish noun a Greek noun has one of three

genders masculine feminine or neuter We can use thisnotion to distinguish nouns that are substantives fromnouns that are adjectives A substantive always keeps thesame gender whereas an adjective agrees with its associ-ated noun in case number and gender (Turkish doesnot show such agreement)

The Greek cases with their rough counterparts inTurkish are as follows

nominative (the dictionary form) genitive (-in hacircli or -den hacircli) dative (-e hacircli or -le hacircli or -de hacircli) accusative (-i hacircli) vocative (usually the same as the nominative and

The stem may be further analyzable as root + characteris-

ticEnglish retains the notion of gender only in its personal pro-

nouns he she it If masculine and feminine are together the an-imate genders and neuter the inanimate then the distinction be-

tween animate and inanimate is shown in whowhich Agreement ofadjective with noun in English is seen in the demonstratives thiswordthese words

One source Oumlzkırımlı [ p ] does indeed treat -le as oneof the durum or hacircl ekleri

anyway it is not needed in mathematics so we shall ignoreit below)

The accusative case is the case of the direct object of averb Turkish assigns the ending -i only to definite directobjects otherwise the nominative is used However for aneuter Greek noun the accusative case is always the sameas the nominative

A Greek noun is of the vowel declension or the conso-nant declension depending on its stem Within the voweldeclension there is a further distinction between the α- orfirst declension and the ο- or second declension Then the

consonant declension is the third declension The spellingof the case of a noun depends on declension and genderTurkish might be said to have four declensions but thevariations in the case-endings in Turkish are determinedby the simple rules of vowel harmony so that it may bemore accurate to say that Turkish has only one declensionSome variations in the Greek endings are due to somethinglike vowel harmony but the rules are much more compli-cated Some examples are in Table

The meanings of the Greek cases are refined by meansof prepositions discussed below

st feminine st feminine nd masculine nd neuter rd neutersingular nominative γραμμή γωνία κύκλος τρίγωνον μέρος

genitive γραμμής γωνίας κύκλου τριγώνου μέρουςdative γραμμῄ γωνίᾳ κύκλῳ τριγώνῳ μέρειaccusative γραμμήν γωνίαν κύκλον τρίγωνον μέρος

plural nominative γραμμαί γωνίαι κύκλοι τρίγωνα μέρηgenitive γραμμών γωνίων κύκλων τριγώνων μέρωνdative γραμμαίς γωνίαις κύκλοις τριγώνοις μέρεσιaccusative γραμμάς γωνίας κύκλους τρίγωνα μέρη

line angle circle triangle part

Table Declension of Greek nouns

The definite article

m f nnom ὁ ἡ τόgen τοῦ τῆς τοῦdat τῷ τῇ τῷacc τόν τήν τό

nom οἱ αἱ τάgen τῶν τῶν τῶνdat τοῖς ταῖς τοῖςacc τούς τάς τά

Table The Greek article

Greek has a definite article corresponding somewhatto the English the Whereas the has only one form theGreek article like an adjective shows distinctions of gen-der number and case with forms as in Table

Euclid may use (a case-form of) τό Α σημεῖον lsquothe Αpointrsquo or ἡ ΑΒ εὐθεία [γραμμή] lsquothe ΑΒ straight [line]rsquoHere the letters Α and ΑΒ come between the article andthe noun in what Smyth calls attributive position [para] Then Α itself is not a point and ΑΒ is not a linethe point and the line are seen in a diagram labelled withthe indicated letters However Euclid may omit the nounspeaking of τό Α lsquothe Αrsquo or ἡ ΑΒ lsquothe ΑΒrsquo

Sometimes (as in Proposition ) a single letter may de-note a straight line but then the letter takes the femininearticle as in ἡ Γ lsquothe Γrsquo since γραμμή lsquolinersquo is feminineNetz [ p] suggests that Euclid uses the neuter

σεμεῖον rather than the feminine στιγμή for lsquopointrsquo so thatpoints and lines will have different genders (See Proposi-tion for a related example)

In general an adjective may be given an article andused as a substantive (Compare lsquoThe best is the enemyof the goodrsquo attributed to Voltaire in the French formLe mieux est lrsquoennemi du bien) The adjective need noteven have the article Euclid usually (but not always) saysstraight instead of straight line and right instead of rightangle In our translation we use straight and rightwhen the substantives straight line and right angle are tobe understood

Euclid may also refer (as in Proposition ) to κοινή ἡΒΓ lsquothe ΒΓ which is commonrsquo Here the adjective κοινήlsquocommonrsquo would appear to be in predicate position [para] In this position the adjective serves not to dis-

English nouns retain a sort of genitive case in the possessiveforms manmanrsquosmenmenrsquos There are further case-distinctionsin pronouns hehishim sheher theytheirthem

httpenwikiquoteorgwikiVoltaire accessed July

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

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Β Γ

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Ε Ζ

Η

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If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

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Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

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The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

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[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

[] Euclid Euclidis Elementa Euclidis Opera Omnia Teubner Edited and with Latin translation by I LHeiberg

[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch

This work is licensed under theCreative Commons

Attribution-NonCommercial-ShareAlike Unported License

To view a copy of this license visithttpcreativecommonsorglicensesby-nc-sa30

or send a letter toCreative Commons

Castro Street Suite Mountain View California USA

Bu ccedilalışmaCreative Commons Attribution-Gayriticari-ShareAlike

Unported Lisansı ile lisanslıLisansın bir kopyasını goumlrebilmek iccedilin

httpcreativecommonsorglicensesby-nc-sa30

adresini ziyaret edin ya da mektup atınCreative Commons

Castro Street Suite Mountain View California USA

CCcopy BYcopy Oumlzer Oumlztuumlrk amp David Pierce $copy

C

copy

Matematik BoumlluumlmuumlMimar Sinan Guumlzel Sanatlar Uumlniversitesi

Bomonti Şişli İstanbul

ozerozturkmsgsuedutr dpiercemsgsuedutr

httpmatmsgsuedutr

This edition of the first book of Euclidrsquos Elements was prepared for a first-year undergraduate course in themathematics department of Mimar Sinan Fine Arts University The text has been corrected after its use in the coursein the fall of

Oumlklidrsquoin Oumlğelerrsquoinin bu baskısı Mimar Sinan Guumlzel Sanatlar Uumlniversitesi Matematik Boumlluumlmnde bir birinci sınıflisans dersi iccedilin hazırlanmıştır ndash Guumlz doumlneminde bu notlar ilk defa kullanılmış ve fark edilen hatalarduumlzeltilmiştir

Introduction

Layout

Book I of Euclidrsquos Elements is presented here in threeparallel columns the original Greek text in the middlecolumn an English translation to its left and a Turkishtranslation to its right

Euclidrsquos Elements consist of books each dividedinto propositions Some books also have definitions

and Book I has also postulates and common notions

In the presentation here the Greek text of each sentenceof each proposition is broken into units so that

each unit will fit on one line the unit as such has a role in the sentence the units kept in the same order make sense when

translated into EnglishEach proposition of the Elements is accompanied by

a picture of points and lines with most points (and somelines) labelled with letters This picture is the lettered

diagram We place the diagram for each proposition af-ter the words According to Reviel Netz [ p n ]this is where the diagram appeared in the original scrollpresumably so that one would know how far to unroll thescroll in order to read the proposition The end of a propo-sition is not to be considered as an undignified positionIndeed Netz judges the diagram to be a metonym forthe proposition something associated with the proposi-tion that is used to stand for the proposition (Today theenunciation of a propositionmdashsee sect belowmdashwould appearto be the common metonym)

Text

We receive Euclidrsquos text through various filters TheElements are supposed to have been composed around bce Heibergrsquos text (published in ) is based mainlyon a manuscript in the Vatican written the tenth centuryce closer to our time than to Euclidrsquos time Knorr []argues that Euclidrsquos original intent may be better reflectedin some Arabic translations from the eighth and ninth cen-turies (The argument is summarized in []) Nonethelesswe shall just use the Heiberg text

More precisely for convenience we take the Greek textin our underlying LATEX file from the LATEX files of RichardFitzpatrick who has published his own parallel Englishtranslation (In the underlying LATEX file the enunci-ation of Proposition I in Greek reads as in Table )Fitzpatrick reports that his Greek text is that of Heiberg

but he gives it without Heibergrsquos apparatus criticus Alsohis method of transcription is unclear There is at leastone mistake in his text (τρὸς for πρὸς near the beginningof I) We shall correct such mistakes if we find themalthough we shall not look for them systematically

In the process of translating we have made use of aprintout of the Greek text of Myungsunn Ryu We donot have a LATEX file for this text only pdf The text issaid to be taken from the Perseus Digital Library

We also refer to images of Heibergrsquos original text []which are available as pdf files from the Wilbour Hallwebsite and from European Cultural Heritage Online(ECHO) In preparing the files from the latter source forprinting we have trimmed the black borders by means ofa program called briss

gtEplsquoi t~hc dojersquoishc egtujersquoiac peperasmrsquoenhc trrsquoigwnon gtisrsquoopleuron sustrsquohsasjai

Table Greek text coded for LATEX

Analysis

Each proposition of the Elements can be understoodas being a problem or a theorem Writing around ce Pappus of Alexandria [ pp ndash] describes the

distinction

Those who favor a more technical terminology ingeometrical research use

httpfarsidephutexasedueuclidhtmlhttpenwikipediaorgwikiFileEuclid-Elementspdfhttpwwwwilbourhallorghttpechompiwg-berlinmpgdehome

httpbrisssourceforgenetIvor Thomas [ p ] uses inquiry here in his translation

but there is no word in the Greek original corresponding to this orto proposition

bull problem (πρόβλημα) to mean a [proposi-tion] in which it is proposed to do or con-struct [something] and

bull theorem (θεώρημα) a [proposition] inwhich the consequences and necessary im-plications of certain hypotheses are investi-gated

but among the ancients some described them allas problems some as theorems

In short a problem proposes something to do a the-orem proposes something to see (The Greek for theoremmeans more generally lsquothat which is looked atrsquo and is re-lated to the verb θεάομαι lsquolook atrsquo from this also comesθέατρον lsquotheaterrsquo)

Be it a problem or a theorem a propositionmdashor moreprecisely the text of a propositionmdashcan be analyzed intoas many as six parts The Green Lion edition [ p xxiii]of Heathrsquos translation of Euclid describes this analysis asfound in Proclusrsquos Commentary on the First Book of Eu-clidrsquos Elements [ p ] In the fifth century ceProclus writes

Every problem and every theorem that is fur-nished with all its parts should contain the fol-lowing elements

) an enunciation (πρότασις)) an exposition (ἔκθεσις)) a specification (διορισμός)) a construction (κατασκευή)) a proof (ἀπόδειξις) and) a conclusion (συμπέρασμα)

Of these the enunciation states what is given andwhat is being sought from it for a perfect enunci-ation consists of both these parts The expositiontakes separately what is given and prepares it inadvance for use in the investigation The specifica-tion takes separately the thing that is sought andmakes clear precisely what it is The construc-tion adds what is lacking in the given for findingwhat is sought The proof draws the proposed in-ference by reasoning scientifically from the propo-sitions that have been admitted The conclusionreverts to the enunciation confirming what hasbeen proved

So many are the parts of a problem or a theoremThe most essential ones and those which are al-ways present are enunciation proof and conclu-sion

Alternative translations are

bull for ἔκθεσις setting out andbull for διορισμός definition of goal [ p ]Heibergrsquos analysis of the text of the Elements into

paragraphs does not correspond exactly to the analysisof Proclus but Netz uses the analysis of Proclus in hisShaping of Deduction in Greek Mathematics [] and weshall use it also according to the following understanding

The enunciation of a proposition is a general state-ment without reference to the lettered diagram Thestatement is about some subject perhaps a straight lineor a triangle

In the exposition that subject is identified in thediagram by means of letters the existence of the subjectis established by means of a third-person imperative verb

(a) The specification of a problem says what will bedone with the subject and it begins with the words δεῖ δὴHere δεῖ is an impersonal verb with the meaning of lsquoit isnecessary torsquo or lsquoit is required torsquo or simply lsquoone mustrsquowhile δή is a lsquotemporal particlersquo with the root meaning oflsquoat this or that pointrsquo [] That which is necessary is ex-pressed by a clause with an infinitive verb In translatingwe may use the English form lsquoIt is necessary for A to beBrsquo(b) The specification of a theorem says what will beproved about the subject and it begins with the wordsλέγω ὅτι lsquoI say thatrsquo The same expression may also ap-pear in a problem in an additional specification at thehead of the proof after the construction

In the construction if it is present the second wordis often γάρ a lsquoconfirmatory adverb and causal conjunc-tionrsquo [ para p ] We translate it as lsquoforrsquo at the be-ginning of the sentence but again γάρ itself is the secondword because it is postpositive it simply never appearsat the beginning of a sentence

Then the proof often begins with the particle ἐπείlsquobecause sincersquo The ἐπεί (or other words) may be fol-lowed by οὖν a lsquoconfirmatory or inferentialrsquo postpositiveparticle [ para p ]

The conclusion repeats the enunciation usuallywith the addition of the postpositive particle ἄρα lsquothere-forersquo Then after the repeated enunciation the conclusionends with one of the clauses(a) ὅπερ ἔδει ποιῆσαι lsquojust what it was necessary to dorsquo(in problems) Heiberg translates this into Latin as quodoportebat fieri although quod erat faciendum or qef isalso used(b) ὅπερ ἔδει δεῖξαι lsquojust what it was necessary to showrsquo (intheorems) in Latin quod erat demonstrandum or qed

Language

The Greek language that we have begun discussing isthe language of Euclid ancient Greek This language be-longs to the so-called Indo-European family of languagesEnglish also belongs to this family but Turkish does not

However in some ways Turkish is closer to Greek thanEnglish is Modern scientific terminology in English orTurkish often has its origins in Greek

Writing

Proclus was born in Byzantium (that is Constantinople nowİstanbul) but his parents were from Lycia (Likya) and he was ed-

ucated first in Xanthus He moved to Alexandria then Athens tostudy philosophy [ p xxxix]

capital minuscule transliteration nameΑ α a alphaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z zetaΗ η ecirc etaΘ θ th thetaΙ ι i iotaΚ κ k kappaΛ λ l lambdaΜ μ m muΝ ν n nuΞ ξ x xiΟ ο o omicronΠ π p piΡ ρ r rhoΣ σv ς s sigmaΤ τ t tauΥ υ y u upsilonΦ φ ph phiΧ χ ch chiΨ ψ ps psiΩ ω ocirc omega

Table The Greek alphabet

The Greek alphabet in Table is the source for theLatin alphabet (which is used by English and Turkish)and it is a source for much scientific symbolism The vow-els of the Greek alphabet are α ε η ι ο υ and ω whereη is a long ε and ω is a long ο the other vowels (α ιυ) can be long or short Some vowels may be given tonalaccents (ά ᾶ ὰ) An initial vowel takes either a rough-breathing mark (as in ἁ) or a smooth-breathing mark (ἀ)the former mark is transliterated by a preceding h andthe latter can be ignored as in ὑπερβολή hyperbolecirc hy-perbola ὀρθογώνιον orthogocircnion rectangle Likewise ῥ istransliterated as rh as in ῥόμβος rhombos rhombus Along vowel may have an iota subscript (ᾳ ῃ ῳ) especially

in case-endings of nouns Of the two forms of minusculesigma the ς appears at the ends of words elsewhere σvappears as in βάσις basis base

In increasing strength the Greek punctuation marksare corresponding to our (The Greekquestion-mark is like our semicolon but it does not ap-pear in Euclid)

Euclid himself will have used only the capital lettersthe minuscules were developed around the ninth century[ para p ] The accent marks were supposedly inventedaround bce because the pronunciation of the ac-cents was dying out [ para p ]

Nouns

As in Turkish so in Greek a single noun or verb canappear in many different forms The general analysis isthe same the noun or verb can be analyzed as stem +ending (goumlvde + ek)

Like a Turkish noun a Greek noun changes to showdistinctions of case and number Unlike a Turkish nouna Greek noun does not take a separate ending (such as-ler) for the plural number rather each case-ending hasa singular form and a plural form (There is also a dualform but this is rarely seen although the distinction be-tween the dual and the plural number occurs for examplein ἑκάτεροςἕκαστος lsquoeithereachrsquo)

Unlike a Turkish noun a Greek noun has one of three

genders masculine feminine or neuter We can use thisnotion to distinguish nouns that are substantives fromnouns that are adjectives A substantive always keeps thesame gender whereas an adjective agrees with its associ-ated noun in case number and gender (Turkish doesnot show such agreement)

The Greek cases with their rough counterparts inTurkish are as follows

nominative (the dictionary form) genitive (-in hacircli or -den hacircli) dative (-e hacircli or -le hacircli or -de hacircli) accusative (-i hacircli) vocative (usually the same as the nominative and

The stem may be further analyzable as root + characteris-

ticEnglish retains the notion of gender only in its personal pro-

nouns he she it If masculine and feminine are together the an-imate genders and neuter the inanimate then the distinction be-

tween animate and inanimate is shown in whowhich Agreement ofadjective with noun in English is seen in the demonstratives thiswordthese words

One source Oumlzkırımlı [ p ] does indeed treat -le as oneof the durum or hacircl ekleri

anyway it is not needed in mathematics so we shall ignoreit below)

The accusative case is the case of the direct object of averb Turkish assigns the ending -i only to definite directobjects otherwise the nominative is used However for aneuter Greek noun the accusative case is always the sameas the nominative

A Greek noun is of the vowel declension or the conso-nant declension depending on its stem Within the voweldeclension there is a further distinction between the α- orfirst declension and the ο- or second declension Then the

consonant declension is the third declension The spellingof the case of a noun depends on declension and genderTurkish might be said to have four declensions but thevariations in the case-endings in Turkish are determinedby the simple rules of vowel harmony so that it may bemore accurate to say that Turkish has only one declensionSome variations in the Greek endings are due to somethinglike vowel harmony but the rules are much more compli-cated Some examples are in Table

The meanings of the Greek cases are refined by meansof prepositions discussed below

st feminine st feminine nd masculine nd neuter rd neutersingular nominative γραμμή γωνία κύκλος τρίγωνον μέρος

genitive γραμμής γωνίας κύκλου τριγώνου μέρουςdative γραμμῄ γωνίᾳ κύκλῳ τριγώνῳ μέρειaccusative γραμμήν γωνίαν κύκλον τρίγωνον μέρος

plural nominative γραμμαί γωνίαι κύκλοι τρίγωνα μέρηgenitive γραμμών γωνίων κύκλων τριγώνων μέρωνdative γραμμαίς γωνίαις κύκλοις τριγώνοις μέρεσιaccusative γραμμάς γωνίας κύκλους τρίγωνα μέρη

line angle circle triangle part

Table Declension of Greek nouns

The definite article

m f nnom ὁ ἡ τόgen τοῦ τῆς τοῦdat τῷ τῇ τῷacc τόν τήν τό

nom οἱ αἱ τάgen τῶν τῶν τῶνdat τοῖς ταῖς τοῖςacc τούς τάς τά

Table The Greek article

Greek has a definite article corresponding somewhatto the English the Whereas the has only one form theGreek article like an adjective shows distinctions of gen-der number and case with forms as in Table

Euclid may use (a case-form of) τό Α σημεῖον lsquothe Αpointrsquo or ἡ ΑΒ εὐθεία [γραμμή] lsquothe ΑΒ straight [line]rsquoHere the letters Α and ΑΒ come between the article andthe noun in what Smyth calls attributive position [para] Then Α itself is not a point and ΑΒ is not a linethe point and the line are seen in a diagram labelled withthe indicated letters However Euclid may omit the nounspeaking of τό Α lsquothe Αrsquo or ἡ ΑΒ lsquothe ΑΒrsquo

Sometimes (as in Proposition ) a single letter may de-note a straight line but then the letter takes the femininearticle as in ἡ Γ lsquothe Γrsquo since γραμμή lsquolinersquo is feminineNetz [ p] suggests that Euclid uses the neuter

σεμεῖον rather than the feminine στιγμή for lsquopointrsquo so thatpoints and lines will have different genders (See Proposi-tion for a related example)

In general an adjective may be given an article andused as a substantive (Compare lsquoThe best is the enemyof the goodrsquo attributed to Voltaire in the French formLe mieux est lrsquoennemi du bien) The adjective need noteven have the article Euclid usually (but not always) saysstraight instead of straight line and right instead of rightangle In our translation we use straight and rightwhen the substantives straight line and right angle are tobe understood

Euclid may also refer (as in Proposition ) to κοινή ἡΒΓ lsquothe ΒΓ which is commonrsquo Here the adjective κοινήlsquocommonrsquo would appear to be in predicate position [para] In this position the adjective serves not to dis-

English nouns retain a sort of genitive case in the possessiveforms manmanrsquosmenmenrsquos There are further case-distinctionsin pronouns hehishim sheher theytheirthem

httpenwikiquoteorgwikiVoltaire accessed July

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

Η

Θ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

Ε

Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

[] Euclid Euclidis Elementa Euclidis Opera Omnia Teubner Edited and with Latin translation by I LHeiberg

[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch

Introduction

Layout

Book I of Euclidrsquos Elements is presented here in threeparallel columns the original Greek text in the middlecolumn an English translation to its left and a Turkishtranslation to its right

Euclidrsquos Elements consist of books each dividedinto propositions Some books also have definitions

and Book I has also postulates and common notions

In the presentation here the Greek text of each sentenceof each proposition is broken into units so that

each unit will fit on one line the unit as such has a role in the sentence the units kept in the same order make sense when

translated into EnglishEach proposition of the Elements is accompanied by

a picture of points and lines with most points (and somelines) labelled with letters This picture is the lettered

diagram We place the diagram for each proposition af-ter the words According to Reviel Netz [ p n ]this is where the diagram appeared in the original scrollpresumably so that one would know how far to unroll thescroll in order to read the proposition The end of a propo-sition is not to be considered as an undignified positionIndeed Netz judges the diagram to be a metonym forthe proposition something associated with the proposi-tion that is used to stand for the proposition (Today theenunciation of a propositionmdashsee sect belowmdashwould appearto be the common metonym)

Text

We receive Euclidrsquos text through various filters TheElements are supposed to have been composed around bce Heibergrsquos text (published in ) is based mainlyon a manuscript in the Vatican written the tenth centuryce closer to our time than to Euclidrsquos time Knorr []argues that Euclidrsquos original intent may be better reflectedin some Arabic translations from the eighth and ninth cen-turies (The argument is summarized in []) Nonethelesswe shall just use the Heiberg text

More precisely for convenience we take the Greek textin our underlying LATEX file from the LATEX files of RichardFitzpatrick who has published his own parallel Englishtranslation (In the underlying LATEX file the enunci-ation of Proposition I in Greek reads as in Table )Fitzpatrick reports that his Greek text is that of Heiberg

but he gives it without Heibergrsquos apparatus criticus Alsohis method of transcription is unclear There is at leastone mistake in his text (τρὸς for πρὸς near the beginningof I) We shall correct such mistakes if we find themalthough we shall not look for them systematically

In the process of translating we have made use of aprintout of the Greek text of Myungsunn Ryu We donot have a LATEX file for this text only pdf The text issaid to be taken from the Perseus Digital Library

We also refer to images of Heibergrsquos original text []which are available as pdf files from the Wilbour Hallwebsite and from European Cultural Heritage Online(ECHO) In preparing the files from the latter source forprinting we have trimmed the black borders by means ofa program called briss

gtEplsquoi t~hc dojersquoishc egtujersquoiac peperasmrsquoenhc trrsquoigwnon gtisrsquoopleuron sustrsquohsasjai

Table Greek text coded for LATEX

Analysis

Each proposition of the Elements can be understoodas being a problem or a theorem Writing around ce Pappus of Alexandria [ pp ndash] describes the

distinction

Those who favor a more technical terminology ingeometrical research use

httpfarsidephutexasedueuclidhtmlhttpenwikipediaorgwikiFileEuclid-Elementspdfhttpwwwwilbourhallorghttpechompiwg-berlinmpgdehome

httpbrisssourceforgenetIvor Thomas [ p ] uses inquiry here in his translation

but there is no word in the Greek original corresponding to this orto proposition

bull problem (πρόβλημα) to mean a [proposi-tion] in which it is proposed to do or con-struct [something] and

bull theorem (θεώρημα) a [proposition] inwhich the consequences and necessary im-plications of certain hypotheses are investi-gated

but among the ancients some described them allas problems some as theorems

In short a problem proposes something to do a the-orem proposes something to see (The Greek for theoremmeans more generally lsquothat which is looked atrsquo and is re-lated to the verb θεάομαι lsquolook atrsquo from this also comesθέατρον lsquotheaterrsquo)

Be it a problem or a theorem a propositionmdashor moreprecisely the text of a propositionmdashcan be analyzed intoas many as six parts The Green Lion edition [ p xxiii]of Heathrsquos translation of Euclid describes this analysis asfound in Proclusrsquos Commentary on the First Book of Eu-clidrsquos Elements [ p ] In the fifth century ceProclus writes

Every problem and every theorem that is fur-nished with all its parts should contain the fol-lowing elements

) an enunciation (πρότασις)) an exposition (ἔκθεσις)) a specification (διορισμός)) a construction (κατασκευή)) a proof (ἀπόδειξις) and) a conclusion (συμπέρασμα)

Of these the enunciation states what is given andwhat is being sought from it for a perfect enunci-ation consists of both these parts The expositiontakes separately what is given and prepares it inadvance for use in the investigation The specifica-tion takes separately the thing that is sought andmakes clear precisely what it is The construc-tion adds what is lacking in the given for findingwhat is sought The proof draws the proposed in-ference by reasoning scientifically from the propo-sitions that have been admitted The conclusionreverts to the enunciation confirming what hasbeen proved

So many are the parts of a problem or a theoremThe most essential ones and those which are al-ways present are enunciation proof and conclu-sion

Alternative translations are

bull for ἔκθεσις setting out andbull for διορισμός definition of goal [ p ]Heibergrsquos analysis of the text of the Elements into

paragraphs does not correspond exactly to the analysisof Proclus but Netz uses the analysis of Proclus in hisShaping of Deduction in Greek Mathematics [] and weshall use it also according to the following understanding

The enunciation of a proposition is a general state-ment without reference to the lettered diagram Thestatement is about some subject perhaps a straight lineor a triangle

In the exposition that subject is identified in thediagram by means of letters the existence of the subjectis established by means of a third-person imperative verb

(a) The specification of a problem says what will bedone with the subject and it begins with the words δεῖ δὴHere δεῖ is an impersonal verb with the meaning of lsquoit isnecessary torsquo or lsquoit is required torsquo or simply lsquoone mustrsquowhile δή is a lsquotemporal particlersquo with the root meaning oflsquoat this or that pointrsquo [] That which is necessary is ex-pressed by a clause with an infinitive verb In translatingwe may use the English form lsquoIt is necessary for A to beBrsquo(b) The specification of a theorem says what will beproved about the subject and it begins with the wordsλέγω ὅτι lsquoI say thatrsquo The same expression may also ap-pear in a problem in an additional specification at thehead of the proof after the construction

In the construction if it is present the second wordis often γάρ a lsquoconfirmatory adverb and causal conjunc-tionrsquo [ para p ] We translate it as lsquoforrsquo at the be-ginning of the sentence but again γάρ itself is the secondword because it is postpositive it simply never appearsat the beginning of a sentence

Then the proof often begins with the particle ἐπείlsquobecause sincersquo The ἐπεί (or other words) may be fol-lowed by οὖν a lsquoconfirmatory or inferentialrsquo postpositiveparticle [ para p ]

The conclusion repeats the enunciation usuallywith the addition of the postpositive particle ἄρα lsquothere-forersquo Then after the repeated enunciation the conclusionends with one of the clauses(a) ὅπερ ἔδει ποιῆσαι lsquojust what it was necessary to dorsquo(in problems) Heiberg translates this into Latin as quodoportebat fieri although quod erat faciendum or qef isalso used(b) ὅπερ ἔδει δεῖξαι lsquojust what it was necessary to showrsquo (intheorems) in Latin quod erat demonstrandum or qed

Language

The Greek language that we have begun discussing isthe language of Euclid ancient Greek This language be-longs to the so-called Indo-European family of languagesEnglish also belongs to this family but Turkish does not

However in some ways Turkish is closer to Greek thanEnglish is Modern scientific terminology in English orTurkish often has its origins in Greek

Writing

Proclus was born in Byzantium (that is Constantinople nowİstanbul) but his parents were from Lycia (Likya) and he was ed-

ucated first in Xanthus He moved to Alexandria then Athens tostudy philosophy [ p xxxix]

capital minuscule transliteration nameΑ α a alphaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z zetaΗ η ecirc etaΘ θ th thetaΙ ι i iotaΚ κ k kappaΛ λ l lambdaΜ μ m muΝ ν n nuΞ ξ x xiΟ ο o omicronΠ π p piΡ ρ r rhoΣ σv ς s sigmaΤ τ t tauΥ υ y u upsilonΦ φ ph phiΧ χ ch chiΨ ψ ps psiΩ ω ocirc omega

Table The Greek alphabet

The Greek alphabet in Table is the source for theLatin alphabet (which is used by English and Turkish)and it is a source for much scientific symbolism The vow-els of the Greek alphabet are α ε η ι ο υ and ω whereη is a long ε and ω is a long ο the other vowels (α ιυ) can be long or short Some vowels may be given tonalaccents (ά ᾶ ὰ) An initial vowel takes either a rough-breathing mark (as in ἁ) or a smooth-breathing mark (ἀ)the former mark is transliterated by a preceding h andthe latter can be ignored as in ὑπερβολή hyperbolecirc hy-perbola ὀρθογώνιον orthogocircnion rectangle Likewise ῥ istransliterated as rh as in ῥόμβος rhombos rhombus Along vowel may have an iota subscript (ᾳ ῃ ῳ) especially

in case-endings of nouns Of the two forms of minusculesigma the ς appears at the ends of words elsewhere σvappears as in βάσις basis base

In increasing strength the Greek punctuation marksare corresponding to our (The Greekquestion-mark is like our semicolon but it does not ap-pear in Euclid)

Euclid himself will have used only the capital lettersthe minuscules were developed around the ninth century[ para p ] The accent marks were supposedly inventedaround bce because the pronunciation of the ac-cents was dying out [ para p ]

Nouns

As in Turkish so in Greek a single noun or verb canappear in many different forms The general analysis isthe same the noun or verb can be analyzed as stem +ending (goumlvde + ek)

Like a Turkish noun a Greek noun changes to showdistinctions of case and number Unlike a Turkish nouna Greek noun does not take a separate ending (such as-ler) for the plural number rather each case-ending hasa singular form and a plural form (There is also a dualform but this is rarely seen although the distinction be-tween the dual and the plural number occurs for examplein ἑκάτεροςἕκαστος lsquoeithereachrsquo)

Unlike a Turkish noun a Greek noun has one of three

genders masculine feminine or neuter We can use thisnotion to distinguish nouns that are substantives fromnouns that are adjectives A substantive always keeps thesame gender whereas an adjective agrees with its associ-ated noun in case number and gender (Turkish doesnot show such agreement)

The Greek cases with their rough counterparts inTurkish are as follows

nominative (the dictionary form) genitive (-in hacircli or -den hacircli) dative (-e hacircli or -le hacircli or -de hacircli) accusative (-i hacircli) vocative (usually the same as the nominative and

The stem may be further analyzable as root + characteris-

ticEnglish retains the notion of gender only in its personal pro-

nouns he she it If masculine and feminine are together the an-imate genders and neuter the inanimate then the distinction be-

tween animate and inanimate is shown in whowhich Agreement ofadjective with noun in English is seen in the demonstratives thiswordthese words

One source Oumlzkırımlı [ p ] does indeed treat -le as oneof the durum or hacircl ekleri

anyway it is not needed in mathematics so we shall ignoreit below)

The accusative case is the case of the direct object of averb Turkish assigns the ending -i only to definite directobjects otherwise the nominative is used However for aneuter Greek noun the accusative case is always the sameas the nominative

A Greek noun is of the vowel declension or the conso-nant declension depending on its stem Within the voweldeclension there is a further distinction between the α- orfirst declension and the ο- or second declension Then the

consonant declension is the third declension The spellingof the case of a noun depends on declension and genderTurkish might be said to have four declensions but thevariations in the case-endings in Turkish are determinedby the simple rules of vowel harmony so that it may bemore accurate to say that Turkish has only one declensionSome variations in the Greek endings are due to somethinglike vowel harmony but the rules are much more compli-cated Some examples are in Table

The meanings of the Greek cases are refined by meansof prepositions discussed below

st feminine st feminine nd masculine nd neuter rd neutersingular nominative γραμμή γωνία κύκλος τρίγωνον μέρος

genitive γραμμής γωνίας κύκλου τριγώνου μέρουςdative γραμμῄ γωνίᾳ κύκλῳ τριγώνῳ μέρειaccusative γραμμήν γωνίαν κύκλον τρίγωνον μέρος

plural nominative γραμμαί γωνίαι κύκλοι τρίγωνα μέρηgenitive γραμμών γωνίων κύκλων τριγώνων μέρωνdative γραμμαίς γωνίαις κύκλοις τριγώνοις μέρεσιaccusative γραμμάς γωνίας κύκλους τρίγωνα μέρη

line angle circle triangle part

Table Declension of Greek nouns

The definite article

m f nnom ὁ ἡ τόgen τοῦ τῆς τοῦdat τῷ τῇ τῷacc τόν τήν τό

nom οἱ αἱ τάgen τῶν τῶν τῶνdat τοῖς ταῖς τοῖςacc τούς τάς τά

Table The Greek article

Greek has a definite article corresponding somewhatto the English the Whereas the has only one form theGreek article like an adjective shows distinctions of gen-der number and case with forms as in Table

Euclid may use (a case-form of) τό Α σημεῖον lsquothe Αpointrsquo or ἡ ΑΒ εὐθεία [γραμμή] lsquothe ΑΒ straight [line]rsquoHere the letters Α and ΑΒ come between the article andthe noun in what Smyth calls attributive position [para] Then Α itself is not a point and ΑΒ is not a linethe point and the line are seen in a diagram labelled withthe indicated letters However Euclid may omit the nounspeaking of τό Α lsquothe Αrsquo or ἡ ΑΒ lsquothe ΑΒrsquo

Sometimes (as in Proposition ) a single letter may de-note a straight line but then the letter takes the femininearticle as in ἡ Γ lsquothe Γrsquo since γραμμή lsquolinersquo is feminineNetz [ p] suggests that Euclid uses the neuter

σεμεῖον rather than the feminine στιγμή for lsquopointrsquo so thatpoints and lines will have different genders (See Proposi-tion for a related example)

In general an adjective may be given an article andused as a substantive (Compare lsquoThe best is the enemyof the goodrsquo attributed to Voltaire in the French formLe mieux est lrsquoennemi du bien) The adjective need noteven have the article Euclid usually (but not always) saysstraight instead of straight line and right instead of rightangle In our translation we use straight and rightwhen the substantives straight line and right angle are tobe understood

Euclid may also refer (as in Proposition ) to κοινή ἡΒΓ lsquothe ΒΓ which is commonrsquo Here the adjective κοινήlsquocommonrsquo would appear to be in predicate position [para] In this position the adjective serves not to dis-

English nouns retain a sort of genitive case in the possessiveforms manmanrsquosmenmenrsquos There are further case-distinctionsin pronouns hehishim sheher theytheirthem

httpenwikiquoteorgwikiVoltaire accessed July

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

Η

Θ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

Ε

Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

[] Euclid Euclidis Elementa Euclidis Opera Omnia Teubner Edited and with Latin translation by I LHeiberg

[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch

bull problem (πρόβλημα) to mean a [proposi-tion] in which it is proposed to do or con-struct [something] and

bull theorem (θεώρημα) a [proposition] inwhich the consequences and necessary im-plications of certain hypotheses are investi-gated

but among the ancients some described them allas problems some as theorems

In short a problem proposes something to do a the-orem proposes something to see (The Greek for theoremmeans more generally lsquothat which is looked atrsquo and is re-lated to the verb θεάομαι lsquolook atrsquo from this also comesθέατρον lsquotheaterrsquo)

Be it a problem or a theorem a propositionmdashor moreprecisely the text of a propositionmdashcan be analyzed intoas many as six parts The Green Lion edition [ p xxiii]of Heathrsquos translation of Euclid describes this analysis asfound in Proclusrsquos Commentary on the First Book of Eu-clidrsquos Elements [ p ] In the fifth century ceProclus writes

Every problem and every theorem that is fur-nished with all its parts should contain the fol-lowing elements

) an enunciation (πρότασις)) an exposition (ἔκθεσις)) a specification (διορισμός)) a construction (κατασκευή)) a proof (ἀπόδειξις) and) a conclusion (συμπέρασμα)

Of these the enunciation states what is given andwhat is being sought from it for a perfect enunci-ation consists of both these parts The expositiontakes separately what is given and prepares it inadvance for use in the investigation The specifica-tion takes separately the thing that is sought andmakes clear precisely what it is The construc-tion adds what is lacking in the given for findingwhat is sought The proof draws the proposed in-ference by reasoning scientifically from the propo-sitions that have been admitted The conclusionreverts to the enunciation confirming what hasbeen proved

So many are the parts of a problem or a theoremThe most essential ones and those which are al-ways present are enunciation proof and conclu-sion

Alternative translations are

bull for ἔκθεσις setting out andbull for διορισμός definition of goal [ p ]Heibergrsquos analysis of the text of the Elements into

paragraphs does not correspond exactly to the analysisof Proclus but Netz uses the analysis of Proclus in hisShaping of Deduction in Greek Mathematics [] and weshall use it also according to the following understanding

The enunciation of a proposition is a general state-ment without reference to the lettered diagram Thestatement is about some subject perhaps a straight lineor a triangle

In the exposition that subject is identified in thediagram by means of letters the existence of the subjectis established by means of a third-person imperative verb

(a) The specification of a problem says what will bedone with the subject and it begins with the words δεῖ δὴHere δεῖ is an impersonal verb with the meaning of lsquoit isnecessary torsquo or lsquoit is required torsquo or simply lsquoone mustrsquowhile δή is a lsquotemporal particlersquo with the root meaning oflsquoat this or that pointrsquo [] That which is necessary is ex-pressed by a clause with an infinitive verb In translatingwe may use the English form lsquoIt is necessary for A to beBrsquo(b) The specification of a theorem says what will beproved about the subject and it begins with the wordsλέγω ὅτι lsquoI say thatrsquo The same expression may also ap-pear in a problem in an additional specification at thehead of the proof after the construction

In the construction if it is present the second wordis often γάρ a lsquoconfirmatory adverb and causal conjunc-tionrsquo [ para p ] We translate it as lsquoforrsquo at the be-ginning of the sentence but again γάρ itself is the secondword because it is postpositive it simply never appearsat the beginning of a sentence

Then the proof often begins with the particle ἐπείlsquobecause sincersquo The ἐπεί (or other words) may be fol-lowed by οὖν a lsquoconfirmatory or inferentialrsquo postpositiveparticle [ para p ]

The conclusion repeats the enunciation usuallywith the addition of the postpositive particle ἄρα lsquothere-forersquo Then after the repeated enunciation the conclusionends with one of the clauses(a) ὅπερ ἔδει ποιῆσαι lsquojust what it was necessary to dorsquo(in problems) Heiberg translates this into Latin as quodoportebat fieri although quod erat faciendum or qef isalso used(b) ὅπερ ἔδει δεῖξαι lsquojust what it was necessary to showrsquo (intheorems) in Latin quod erat demonstrandum or qed

Language

The Greek language that we have begun discussing isthe language of Euclid ancient Greek This language be-longs to the so-called Indo-European family of languagesEnglish also belongs to this family but Turkish does not

However in some ways Turkish is closer to Greek thanEnglish is Modern scientific terminology in English orTurkish often has its origins in Greek

Writing

Proclus was born in Byzantium (that is Constantinople nowİstanbul) but his parents were from Lycia (Likya) and he was ed-

ucated first in Xanthus He moved to Alexandria then Athens tostudy philosophy [ p xxxix]

capital minuscule transliteration nameΑ α a alphaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z zetaΗ η ecirc etaΘ θ th thetaΙ ι i iotaΚ κ k kappaΛ λ l lambdaΜ μ m muΝ ν n nuΞ ξ x xiΟ ο o omicronΠ π p piΡ ρ r rhoΣ σv ς s sigmaΤ τ t tauΥ υ y u upsilonΦ φ ph phiΧ χ ch chiΨ ψ ps psiΩ ω ocirc omega

Table The Greek alphabet

The Greek alphabet in Table is the source for theLatin alphabet (which is used by English and Turkish)and it is a source for much scientific symbolism The vow-els of the Greek alphabet are α ε η ι ο υ and ω whereη is a long ε and ω is a long ο the other vowels (α ιυ) can be long or short Some vowels may be given tonalaccents (ά ᾶ ὰ) An initial vowel takes either a rough-breathing mark (as in ἁ) or a smooth-breathing mark (ἀ)the former mark is transliterated by a preceding h andthe latter can be ignored as in ὑπερβολή hyperbolecirc hy-perbola ὀρθογώνιον orthogocircnion rectangle Likewise ῥ istransliterated as rh as in ῥόμβος rhombos rhombus Along vowel may have an iota subscript (ᾳ ῃ ῳ) especially

in case-endings of nouns Of the two forms of minusculesigma the ς appears at the ends of words elsewhere σvappears as in βάσις basis base

In increasing strength the Greek punctuation marksare corresponding to our (The Greekquestion-mark is like our semicolon but it does not ap-pear in Euclid)

Euclid himself will have used only the capital lettersthe minuscules were developed around the ninth century[ para p ] The accent marks were supposedly inventedaround bce because the pronunciation of the ac-cents was dying out [ para p ]

Nouns

As in Turkish so in Greek a single noun or verb canappear in many different forms The general analysis isthe same the noun or verb can be analyzed as stem +ending (goumlvde + ek)

Like a Turkish noun a Greek noun changes to showdistinctions of case and number Unlike a Turkish nouna Greek noun does not take a separate ending (such as-ler) for the plural number rather each case-ending hasa singular form and a plural form (There is also a dualform but this is rarely seen although the distinction be-tween the dual and the plural number occurs for examplein ἑκάτεροςἕκαστος lsquoeithereachrsquo)

Unlike a Turkish noun a Greek noun has one of three

genders masculine feminine or neuter We can use thisnotion to distinguish nouns that are substantives fromnouns that are adjectives A substantive always keeps thesame gender whereas an adjective agrees with its associ-ated noun in case number and gender (Turkish doesnot show such agreement)

The Greek cases with their rough counterparts inTurkish are as follows

nominative (the dictionary form) genitive (-in hacircli or -den hacircli) dative (-e hacircli or -le hacircli or -de hacircli) accusative (-i hacircli) vocative (usually the same as the nominative and

The stem may be further analyzable as root + characteris-

ticEnglish retains the notion of gender only in its personal pro-

nouns he she it If masculine and feminine are together the an-imate genders and neuter the inanimate then the distinction be-

tween animate and inanimate is shown in whowhich Agreement ofadjective with noun in English is seen in the demonstratives thiswordthese words

One source Oumlzkırımlı [ p ] does indeed treat -le as oneof the durum or hacircl ekleri

anyway it is not needed in mathematics so we shall ignoreit below)

The accusative case is the case of the direct object of averb Turkish assigns the ending -i only to definite directobjects otherwise the nominative is used However for aneuter Greek noun the accusative case is always the sameas the nominative

A Greek noun is of the vowel declension or the conso-nant declension depending on its stem Within the voweldeclension there is a further distinction between the α- orfirst declension and the ο- or second declension Then the

consonant declension is the third declension The spellingof the case of a noun depends on declension and genderTurkish might be said to have four declensions but thevariations in the case-endings in Turkish are determinedby the simple rules of vowel harmony so that it may bemore accurate to say that Turkish has only one declensionSome variations in the Greek endings are due to somethinglike vowel harmony but the rules are much more compli-cated Some examples are in Table

The meanings of the Greek cases are refined by meansof prepositions discussed below

st feminine st feminine nd masculine nd neuter rd neutersingular nominative γραμμή γωνία κύκλος τρίγωνον μέρος

genitive γραμμής γωνίας κύκλου τριγώνου μέρουςdative γραμμῄ γωνίᾳ κύκλῳ τριγώνῳ μέρειaccusative γραμμήν γωνίαν κύκλον τρίγωνον μέρος

plural nominative γραμμαί γωνίαι κύκλοι τρίγωνα μέρηgenitive γραμμών γωνίων κύκλων τριγώνων μέρωνdative γραμμαίς γωνίαις κύκλοις τριγώνοις μέρεσιaccusative γραμμάς γωνίας κύκλους τρίγωνα μέρη

line angle circle triangle part

Table Declension of Greek nouns

The definite article

m f nnom ὁ ἡ τόgen τοῦ τῆς τοῦdat τῷ τῇ τῷacc τόν τήν τό

nom οἱ αἱ τάgen τῶν τῶν τῶνdat τοῖς ταῖς τοῖςacc τούς τάς τά

Table The Greek article

Greek has a definite article corresponding somewhatto the English the Whereas the has only one form theGreek article like an adjective shows distinctions of gen-der number and case with forms as in Table

Euclid may use (a case-form of) τό Α σημεῖον lsquothe Αpointrsquo or ἡ ΑΒ εὐθεία [γραμμή] lsquothe ΑΒ straight [line]rsquoHere the letters Α and ΑΒ come between the article andthe noun in what Smyth calls attributive position [para] Then Α itself is not a point and ΑΒ is not a linethe point and the line are seen in a diagram labelled withthe indicated letters However Euclid may omit the nounspeaking of τό Α lsquothe Αrsquo or ἡ ΑΒ lsquothe ΑΒrsquo

Sometimes (as in Proposition ) a single letter may de-note a straight line but then the letter takes the femininearticle as in ἡ Γ lsquothe Γrsquo since γραμμή lsquolinersquo is feminineNetz [ p] suggests that Euclid uses the neuter

σεμεῖον rather than the feminine στιγμή for lsquopointrsquo so thatpoints and lines will have different genders (See Proposi-tion for a related example)

In general an adjective may be given an article andused as a substantive (Compare lsquoThe best is the enemyof the goodrsquo attributed to Voltaire in the French formLe mieux est lrsquoennemi du bien) The adjective need noteven have the article Euclid usually (but not always) saysstraight instead of straight line and right instead of rightangle In our translation we use straight and rightwhen the substantives straight line and right angle are tobe understood

Euclid may also refer (as in Proposition ) to κοινή ἡΒΓ lsquothe ΒΓ which is commonrsquo Here the adjective κοινήlsquocommonrsquo would appear to be in predicate position [para] In this position the adjective serves not to dis-

English nouns retain a sort of genitive case in the possessiveforms manmanrsquosmenmenrsquos There are further case-distinctionsin pronouns hehishim sheher theytheirthem

httpenwikiquoteorgwikiVoltaire accessed July

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

Η

Θ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

Ε

Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

[] Euclid Euclidis Elementa Euclidis Opera Omnia Teubner Edited and with Latin translation by I LHeiberg

[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch

capital minuscule transliteration nameΑ α a alphaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z zetaΗ η ecirc etaΘ θ th thetaΙ ι i iotaΚ κ k kappaΛ λ l lambdaΜ μ m muΝ ν n nuΞ ξ x xiΟ ο o omicronΠ π p piΡ ρ r rhoΣ σv ς s sigmaΤ τ t tauΥ υ y u upsilonΦ φ ph phiΧ χ ch chiΨ ψ ps psiΩ ω ocirc omega

Table The Greek alphabet

The Greek alphabet in Table is the source for theLatin alphabet (which is used by English and Turkish)and it is a source for much scientific symbolism The vow-els of the Greek alphabet are α ε η ι ο υ and ω whereη is a long ε and ω is a long ο the other vowels (α ιυ) can be long or short Some vowels may be given tonalaccents (ά ᾶ ὰ) An initial vowel takes either a rough-breathing mark (as in ἁ) or a smooth-breathing mark (ἀ)the former mark is transliterated by a preceding h andthe latter can be ignored as in ὑπερβολή hyperbolecirc hy-perbola ὀρθογώνιον orthogocircnion rectangle Likewise ῥ istransliterated as rh as in ῥόμβος rhombos rhombus Along vowel may have an iota subscript (ᾳ ῃ ῳ) especially

in case-endings of nouns Of the two forms of minusculesigma the ς appears at the ends of words elsewhere σvappears as in βάσις basis base

In increasing strength the Greek punctuation marksare corresponding to our (The Greekquestion-mark is like our semicolon but it does not ap-pear in Euclid)

Euclid himself will have used only the capital lettersthe minuscules were developed around the ninth century[ para p ] The accent marks were supposedly inventedaround bce because the pronunciation of the ac-cents was dying out [ para p ]

Nouns

As in Turkish so in Greek a single noun or verb canappear in many different forms The general analysis isthe same the noun or verb can be analyzed as stem +ending (goumlvde + ek)

Like a Turkish noun a Greek noun changes to showdistinctions of case and number Unlike a Turkish nouna Greek noun does not take a separate ending (such as-ler) for the plural number rather each case-ending hasa singular form and a plural form (There is also a dualform but this is rarely seen although the distinction be-tween the dual and the plural number occurs for examplein ἑκάτεροςἕκαστος lsquoeithereachrsquo)

Unlike a Turkish noun a Greek noun has one of three

genders masculine feminine or neuter We can use thisnotion to distinguish nouns that are substantives fromnouns that are adjectives A substantive always keeps thesame gender whereas an adjective agrees with its associ-ated noun in case number and gender (Turkish doesnot show such agreement)

The Greek cases with their rough counterparts inTurkish are as follows

nominative (the dictionary form) genitive (-in hacircli or -den hacircli) dative (-e hacircli or -le hacircli or -de hacircli) accusative (-i hacircli) vocative (usually the same as the nominative and

The stem may be further analyzable as root + characteris-

ticEnglish retains the notion of gender only in its personal pro-

nouns he she it If masculine and feminine are together the an-imate genders and neuter the inanimate then the distinction be-

tween animate and inanimate is shown in whowhich Agreement ofadjective with noun in English is seen in the demonstratives thiswordthese words

One source Oumlzkırımlı [ p ] does indeed treat -le as oneof the durum or hacircl ekleri

anyway it is not needed in mathematics so we shall ignoreit below)

The accusative case is the case of the direct object of averb Turkish assigns the ending -i only to definite directobjects otherwise the nominative is used However for aneuter Greek noun the accusative case is always the sameas the nominative

A Greek noun is of the vowel declension or the conso-nant declension depending on its stem Within the voweldeclension there is a further distinction between the α- orfirst declension and the ο- or second declension Then the

consonant declension is the third declension The spellingof the case of a noun depends on declension and genderTurkish might be said to have four declensions but thevariations in the case-endings in Turkish are determinedby the simple rules of vowel harmony so that it may bemore accurate to say that Turkish has only one declensionSome variations in the Greek endings are due to somethinglike vowel harmony but the rules are much more compli-cated Some examples are in Table

The meanings of the Greek cases are refined by meansof prepositions discussed below

st feminine st feminine nd masculine nd neuter rd neutersingular nominative γραμμή γωνία κύκλος τρίγωνον μέρος

genitive γραμμής γωνίας κύκλου τριγώνου μέρουςdative γραμμῄ γωνίᾳ κύκλῳ τριγώνῳ μέρειaccusative γραμμήν γωνίαν κύκλον τρίγωνον μέρος

plural nominative γραμμαί γωνίαι κύκλοι τρίγωνα μέρηgenitive γραμμών γωνίων κύκλων τριγώνων μέρωνdative γραμμαίς γωνίαις κύκλοις τριγώνοις μέρεσιaccusative γραμμάς γωνίας κύκλους τρίγωνα μέρη

line angle circle triangle part

Table Declension of Greek nouns

The definite article

m f nnom ὁ ἡ τόgen τοῦ τῆς τοῦdat τῷ τῇ τῷacc τόν τήν τό

nom οἱ αἱ τάgen τῶν τῶν τῶνdat τοῖς ταῖς τοῖςacc τούς τάς τά

Table The Greek article

Greek has a definite article corresponding somewhatto the English the Whereas the has only one form theGreek article like an adjective shows distinctions of gen-der number and case with forms as in Table

Euclid may use (a case-form of) τό Α σημεῖον lsquothe Αpointrsquo or ἡ ΑΒ εὐθεία [γραμμή] lsquothe ΑΒ straight [line]rsquoHere the letters Α and ΑΒ come between the article andthe noun in what Smyth calls attributive position [para] Then Α itself is not a point and ΑΒ is not a linethe point and the line are seen in a diagram labelled withthe indicated letters However Euclid may omit the nounspeaking of τό Α lsquothe Αrsquo or ἡ ΑΒ lsquothe ΑΒrsquo

Sometimes (as in Proposition ) a single letter may de-note a straight line but then the letter takes the femininearticle as in ἡ Γ lsquothe Γrsquo since γραμμή lsquolinersquo is feminineNetz [ p] suggests that Euclid uses the neuter

σεμεῖον rather than the feminine στιγμή for lsquopointrsquo so thatpoints and lines will have different genders (See Proposi-tion for a related example)

In general an adjective may be given an article andused as a substantive (Compare lsquoThe best is the enemyof the goodrsquo attributed to Voltaire in the French formLe mieux est lrsquoennemi du bien) The adjective need noteven have the article Euclid usually (but not always) saysstraight instead of straight line and right instead of rightangle In our translation we use straight and rightwhen the substantives straight line and right angle are tobe understood

Euclid may also refer (as in Proposition ) to κοινή ἡΒΓ lsquothe ΒΓ which is commonrsquo Here the adjective κοινήlsquocommonrsquo would appear to be in predicate position [para] In this position the adjective serves not to dis-

English nouns retain a sort of genitive case in the possessiveforms manmanrsquosmenmenrsquos There are further case-distinctionsin pronouns hehishim sheher theytheirthem

httpenwikiquoteorgwikiVoltaire accessed July

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

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Β Γ

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Ε Ζ

Η

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If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

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Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

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The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

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[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

[] Euclid Euclidis Elementa Euclidis Opera Omnia Teubner Edited and with Latin translation by I LHeiberg

[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch

anyway it is not needed in mathematics so we shall ignoreit below)

The accusative case is the case of the direct object of averb Turkish assigns the ending -i only to definite directobjects otherwise the nominative is used However for aneuter Greek noun the accusative case is always the sameas the nominative

A Greek noun is of the vowel declension or the conso-nant declension depending on its stem Within the voweldeclension there is a further distinction between the α- orfirst declension and the ο- or second declension Then the

consonant declension is the third declension The spellingof the case of a noun depends on declension and genderTurkish might be said to have four declensions but thevariations in the case-endings in Turkish are determinedby the simple rules of vowel harmony so that it may bemore accurate to say that Turkish has only one declensionSome variations in the Greek endings are due to somethinglike vowel harmony but the rules are much more compli-cated Some examples are in Table

The meanings of the Greek cases are refined by meansof prepositions discussed below

st feminine st feminine nd masculine nd neuter rd neutersingular nominative γραμμή γωνία κύκλος τρίγωνον μέρος

genitive γραμμής γωνίας κύκλου τριγώνου μέρουςdative γραμμῄ γωνίᾳ κύκλῳ τριγώνῳ μέρειaccusative γραμμήν γωνίαν κύκλον τρίγωνον μέρος

plural nominative γραμμαί γωνίαι κύκλοι τρίγωνα μέρηgenitive γραμμών γωνίων κύκλων τριγώνων μέρωνdative γραμμαίς γωνίαις κύκλοις τριγώνοις μέρεσιaccusative γραμμάς γωνίας κύκλους τρίγωνα μέρη

line angle circle triangle part

Table Declension of Greek nouns

The definite article

m f nnom ὁ ἡ τόgen τοῦ τῆς τοῦdat τῷ τῇ τῷacc τόν τήν τό

nom οἱ αἱ τάgen τῶν τῶν τῶνdat τοῖς ταῖς τοῖςacc τούς τάς τά

Table The Greek article

Greek has a definite article corresponding somewhatto the English the Whereas the has only one form theGreek article like an adjective shows distinctions of gen-der number and case with forms as in Table

Euclid may use (a case-form of) τό Α σημεῖον lsquothe Αpointrsquo or ἡ ΑΒ εὐθεία [γραμμή] lsquothe ΑΒ straight [line]rsquoHere the letters Α and ΑΒ come between the article andthe noun in what Smyth calls attributive position [para] Then Α itself is not a point and ΑΒ is not a linethe point and the line are seen in a diagram labelled withthe indicated letters However Euclid may omit the nounspeaking of τό Α lsquothe Αrsquo or ἡ ΑΒ lsquothe ΑΒrsquo

Sometimes (as in Proposition ) a single letter may de-note a straight line but then the letter takes the femininearticle as in ἡ Γ lsquothe Γrsquo since γραμμή lsquolinersquo is feminineNetz [ p] suggests that Euclid uses the neuter

σεμεῖον rather than the feminine στιγμή for lsquopointrsquo so thatpoints and lines will have different genders (See Proposi-tion for a related example)

In general an adjective may be given an article andused as a substantive (Compare lsquoThe best is the enemyof the goodrsquo attributed to Voltaire in the French formLe mieux est lrsquoennemi du bien) The adjective need noteven have the article Euclid usually (but not always) saysstraight instead of straight line and right instead of rightangle In our translation we use straight and rightwhen the substantives straight line and right angle are tobe understood

Euclid may also refer (as in Proposition ) to κοινή ἡΒΓ lsquothe ΒΓ which is commonrsquo Here the adjective κοινήlsquocommonrsquo would appear to be in predicate position [para] In this position the adjective serves not to dis-

English nouns retain a sort of genitive case in the possessiveforms manmanrsquosmenmenrsquos There are further case-distinctionsin pronouns hehishim sheher theytheirthem

httpenwikiquoteorgwikiVoltaire accessed July

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

Η

Θ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

Ε

Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

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[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch

tinguish the straight line in question from other straightlines but to express its relation to other parts of the di-agram (in this case that it is the base of two differenttriangles)

Similarly Euclid may use the adjective ὅλος wholein predicate position as in Proposition ὅλον τὸ ΑΒΓτρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει lsquothe ΑΒΓtriangle as a whole to the ΔΕΖ triangle as a whole willapplyrsquo Smythrsquos examples of adjective position includeattributive τὸ ὅλον στράτευμα the whole armypredicate ὅλον τὸ στράτευμα the army as a wholeThe distinction here may be that the whole army may haveattributes of a person as in lsquoThe whole army is hungryrsquobut the army as a whole does not (as a whole it is nota person) The distinction is subtle and in the example

from Euclid Heath just gives the translation lsquothe wholetrianglersquo

In Proposition Euclid refers to ἡ ὑπὸ ΑΒΓ γωνίαwhich perhaps stands for ἡ περιεχομένη ὑπὸ τῆς ΑΒΓγραμμὴς γωνία lsquothe contained-by-the-ΑΒΓ-line anglersquo orἡ περιεχομένη ὑπὸ τῶν ΑΒ ΒΓ ευθείων γραμμὼν γωνίαlsquothe bounded-by-the-ΑΒ-ΒΓ-straight-lines anglersquo In thesame proposition the form γωνία ἡ ὑπὸ ΑΒΓ appears (ac-tually γωνία ἡ ὑπὸ ΒΖΓ) with no obvious distinction inmeaning (Each position of [ἡ] ὑπὸ ΑΒΓ is called attribu-tive by Smyth) For short Euclid may say just ἡ ὑπὸ ΑΒΓfor the angle without using γωνία

The nesting of adjectives between article and noun canbe repeated An extreme example is the phrase from theenunciation of Proposition analyzed in Table

τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνονἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς

τὴν ὀρθὴν γωνίαν

the right angleon the side subtending the right angle

the square on the side subtending the right angle

Table Nesting of Greek adjective phrases

Prepositions

In the example in Table the preposition ἀπό appearsThis is used only before nouns in the genitive case It usu-ally has the sense of the English preposition from as inthe first postulate or in the construction of Proposition where straight lines are drawn from the point Γ to Α andΒ In Table then the sense of the Greek is not exactlythat the square sits on the side but that it arises from theside

Euclid uses various prepositions which when used be-fore nouns in various cases have meanings roughly as inTable Details follow

When its object is in the accusative case the prepo-sition ἐπί has the sense of the English preposition to asagain in in the first postulate or in the construction ofProposition where straight lines are drawn from Γ to Αand Β

The prepositional phrase ἐπὶ τὰ αὐτὰ μέρη lsquoto the samepartsrsquo is used several times as for example in the fifthpostulate and Proposition The object of the preposi-tion ἐπί is again in the accusative case but is plural Itwould appear that as in English so in Greek lsquopartsrsquo canhave the sense of the singular lsquoregionrsquo More precisely inthis case the meaning of lsquopartsrsquo would appear to be lsquoside[of a straight line]rsquo and one might translate the phrase ἐπὶτὰ αὐτὰ μέρη by lsquoon the same sidersquo (as Heath does) Themore general sense of lsquopartrsquo is used in the fifth commonnotion

The object of the preposition ἐπί may also be in the

genitive case Then ἐπί has the sense of on as yet againin the construction of Proposition where a triangle isconstructed on the straight line ΑΒ

The preposition πρός is used in the set phrase πρὸςὀρθὰς [γωνίας] at right angles where the noun phrase ὀρθὴ[γωνία] right [angle] is a plural accusative Also in thedefinitions of angle and circle πρός is used with the ac-cusative in a sense normally expressed in English by lsquotorsquoIn every other case in Euclidrsquos Book I πρός is used withthe dative case and also has the sense of at or on as forexample in Proposition where a straight line is to beplaced at a given point

There is a set phrase used in Propositions and in which πρός appears twice πρὸςτῇ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ lsquoat the straight [line]and [at ] the point on itrsquo (It is assumed here that thefirst occurrence of πρός takes two objects both straightand point It is unlikely that point is ungoverned sinceaccording to Smyth [ para] in prose lsquothe dative ofplace (chiefly place where) is used only of proper namesrsquo)

The preposition διά is used with the accusative caseto give explanations The explanation might be a clausewhose verb is an infinitive and whose subject is in theaccusative case itself then the whole clause is given theaccusative case by being preceded by the neuter accusativearticle τό The first example is in Proposition διὰ τὸἴσην εἶναι τὴν ΑΒ τῇ ΔΕ lsquobecause ΑΒ is equal to ΔΕrsquo

The preposition διά is also used with the genitive caseThis is an elaboration of an observation by Netz [ p

pp -]According to Netz [ p ] lsquopartsrsquo means lsquodirectionrsquo in

this phrase and only in this phrase

It may however be pointed out that the article τό could also bein the nominative case However prepositions are never followed bya case that is unambiguously nominative

with the sense of through as in speaking of a straight linethrough a point This use of διά always occurs in a setphrase as in the enunciation of Proposition where thestraight line through the point is also parallel to someother straight line

The preposition κατά is used in Book I always with aname or a word for a point in the accusative case Thispoint may be where two straight lines meet as in Proposi-tion or where a straight line is bisected as in Proposi-tion The set phrase κατὰ κορυφήν lsquoat a headrsquo occurs forexample in the enunciation of Proposition to describeangles that are lsquovertically oppositersquo or simply vertical

The preposition μετά used with the genitive casemeans with It occurs in Book I only in Proposition only with the names of triangles only in the sentence τὸΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳμετὰ τοῦ ΚΖΓ lsquoTriangle ΑΕΚ with [triangle] ΚΗΓ is equalto triangle ΑΘΚ with [triangle] ΚΖΓrsquo

The preposition παρά is used in Book I only in Propo-sition with the name of a straight line in the genitivecase and then the preposition has the sense of along aparallelogram is to be constructed one of whose sides isset along the original straight line so that they coincide

The adjective παράλληλος lsquoparallelrsquo used frequentlystarting with Proposition seems to result from παρά+ ἀλλήλων lsquoalongside one anotherrsquo Here ἀλλήλων is thereciprocal pronoun lsquoone anotherrsquo never used in the singu-lar or nominative it seems to result from ἀ΄λλος lsquoanotherrsquoThe dative plural ἀλλήλοις occurs frequently as in Propo-sition where circles cut one another and two straightlines are equal to one another

The preposition ὑπό is used in naming angles by let-ters as in ἡ ὑπὸ ΑΒΓ γωνία lsquothe angle ΑΒΓrsquo Possibly sucha phrase arises from a longer phrase as in Proposition ἡ γωνία ἡ ὑπὸ τῶν εὐθειῶν περιεχομένη lsquothe angle that is

contained by the [two] sides [elsewhere indicated]rsquo Hereὑπό precedes the agent of a passive verb and the noun forthe agent is in the genitive case There is a similar use inthe enunciation of Proposition ἡ ὑπὸ ΒΑΓ γωνία δίχατέτμηται ὑπὸ τῆς ΑΖ εὐθείας lsquoThe angle ΒΑΓ is bisectedby the [straight line] ΑΖrsquo

The preposition ὑπό is also used with nouns in the ac-cusative case It may then have the meaning of underas in Proposition More commonly it just precedes ob-jects of the verb ὑποτείνω lsquostretch underrsquo used in Englishin the Latinate form subtend The subject of this verbwill be the side of a triangle and the object will be theopposite angle

The preposition ἐν lsquoinrsquo is used only with the dativefrequently in the phrase ἐν ταῖς αὐταῖς παραλλήλοις lsquoin thesame parallelsrsquo starting with Proposition It is used inProposition and later with reference to parallelogramsin a given angle Finally in Proposition (the so-calledPythagorean Theorem) there is a general reference to asituation in right-angled triangles

The preposition ἐξ lsquofromrsquo is used with the genitive caseIn Proposition in the set phrase ἐξ αρχῆς lsquofrom the be-ginningrsquo that is original Beyond this ἐξ appears onlyin the problematic definitions of straight line and planesurface in the set phrase ἐξ ἰσού lsquofrom equalityrsquo or asHeath has it lsquoevenlyrsquo

The preposition περί lsquoaboutrsquo is used only in Proposi-tions and only with the accusative only with refer-ence to figures arranged about the diameter of a parallel-ogram

Greek has a few other prepositions σύν ἀντί πρόἀμφί and ὑπέρ but these are not used in Book I Any ofthe prepositions may be used also as a prefix in a noun orverb

Verbs

A verb may show distinctions of person number voicetense mood (mode) and aspect Names for the forms thatoccur in Euclid are

mood indicative imperative or subjunctive

aspect continuous perfect or aorist

number singular or plural

voice active or passive

person first or third

tense past present or future

(In other Greek writing there are also a second persona dual number and an optative mood One speaks of amiddle voice but this usually has the same form as thepassive) Euclid also uses verbal nouns namely infinitives(verbal substantives) and participles (verbal adjectives)

Suppose the utterance of a sentence involves threethings the speaker of the sentence the act described by

the sentence and the performer of the act If only for thesake of remembering the six verb features above one canmake associations as follows

mood speaker aspect act number performer voice performerndashact person speakerndashperformer tense actndashspeakerFirst-person verbs are rare in Euclid As noted above

λέγω lsquoI sayrsquo is used at the beginning of specifications oftheorems and a few other places Also δείξομεν lsquowe shallshowrsquo is used a few times The other verbs are in the thirdperson

Of the propositions of Book I have enunciationsof the form ᾿Εάν + subjunctive

Often in sentences of the logical form lsquoIf A then BrsquoEuclid will express lsquoIf Arsquo as a genitive absolute a noun andparticiple in the genitive case We use the correspondingabsolute construction in English

Translation

genitive dative accusativeἀπό fromδιά through [a point] owing toἐν inἐξ from [the beginning]ἐπι on toκατά at [a point]μετά withπαρά along [a straight line]περί aboutπρός aton at [right angles]ὑπό by under

Table Greek prepositions

The Perseus website with its Word Study Tool isuseful for parsing However in the work of interpret-ing the Greek we also consult print resources such asSmythrsquos Greek Grammar [] the Greek-English Lexiconof Liddell Scott and Jones [] the Pocket Oxford Clas-sical Greek Dictionary [] and Heathrsquos translation of theElements [ ]

There are online lessons on reading Euclid in Greek

In translating Euclid into English Heath seems to stayas close to Euclid as possible under the requirement thatthe translation still read well as English There may besubtle ways in which Heath imposes modern ways of think-ing that are foreign to Euclid

The English translation here tries to stay even closerto Euclid than Heath does The purpose of the transla-tion is to elucidate the original Greek This means thetranslation may not read so well as English In partic-ular word order may be odd Simple declarative sen-tences in English normally have the order subject-verb-object (or subject-copula-predicate) When Eu-clid uses another order say subject-object-verb (orsubject-predicate-copula) the translation may fol-low him There is a precedent for such variations in En-glish order albeit from a few centuries ago For examplethere is the rendition by George Chapman (ndash) ofHomerrsquos Iliad [] Chapman begins his version of Homerthus

Achillesrsquo banefull wrath resound O Goddessethat imposd

Infinite sorrowes on the Greekes and manybrave soules losd

From breasts Heroiquemdashsent them farre tothat invisible cave

That no light comforts and their lims to dogsand vultures gave

To all which Joversquos will gave effect from whomfirst strife begunne

Betwixt Atrides king of men and Thetisrsquo god-

like Sonne

The word order subject-predicate-copula is seenalso in the lines of Sir Walter Raleigh (ndash)quoted approvingly by Henry David Thoreau (ndash) []

But men labor under a mistake The better partof the man is soon plowed into the soil for com-post By a seeming fate commonly called neces-sity they are employed as it says in an old booklaying up treasures which moth and rust will cor-rupt and thieves break through and steal It isa foolrsquos life as they will find when they get to theend of it if not before It is said that Deucalionand Pyrrha created men by throwing stones overtheir heads behind themmdash

ldquoInde genus durum sumus experien-

sque laborum

Et documenta damus qua simus origine

natirdquo

Or as Raleigh rhymes it in his sonorous waymdash

ldquoFrom thence our kind hard-hearted is

enduring pain and care

Approving that our bodies of a stony

nature arerdquo

So much for a blind obedience to a blundering or-acle throwing the stones over their heads behindthem and not seeing where they fell

More examples

The man recovered of the biteThe dog it was that died

Whose woods these are I think I knowHis house is in the village thoughHe will not see me stopping hereTo watch his woods fill up with snow

httpwwwperseustuftseduhoppercollection

collection=Perseus3Acorpus3Aperseus2Cwork2CEuclid

2C20Elementshttpwwwduedu~etuttleclassicsnugreekcontents

htmThe Gospel According to St Matthew lsquoLay not up for

yourselves treasures upon earth where moth and rust doth corruptand where thieves break through and stealrsquo

Text taken from httpwwwgutenbergorgfiles205205-h

205-hhtm July The last lines of lsquoAn Elegy on the Death of a Mad Dogrsquo by

Oliver Goldsmith (-) (httpwwwpoetry-archivecomgan_elegy_on_the_death_of_a_mad_doghtml accessed July )

The first stanza of lsquoStopping by Woods on a Snowy Eveningrsquoby Robert Frost (httpwwwpoetryfoundationorgpoem171621accessed July )

Giriş

Sayfa duumlzeni ve Metin

Oumlklidrsquoin Oumlğeler inin birinci kitabı burada uumlccedil suumltunhalinde sunuluyor orta suumltunda orijinal Yunanca metinonun solunda bir İngilizce ccedilevirisi ve sağında bir Tuumlrkccedileccedilevirisi yer alıyor

Oumlklidrsquoin Oumlğeler i her biri oumlnermelere boumlluumlnmuumlş olan kitaptan oluşur Bazı kitaplarda tanımlar da vardırBirinci kitap ayrıca postuumllatları ve genel kavramları

da iccedilerir Yunanca metnin her oumlnermesinin her cuumlmlesioumlyle birimlere boumlluumlnmuumlştuumlr ki

her birim bir satıra sığar birimler cuumlmle iccedilinde bir rol oynarlar İngilizceye ccedilevirirken birimlerin sırasını korumak an-

lamlı olur

Oumlğeler in her oumlnermesinin yanında ccediloğu noktanın (vebazı ccedilizgilerin) harflerle isimlendirildiği bir ccedilizgi ve nokta-lar resmi yer alır Bu resim harfli diagramdır Her oumlner-mede diagramı kelimelerin sonuna yerleştiriyoruz RevielNetzrsquoe goumlre orijinal ruloda diagram burada yer alırdı veboumlylece okuyan oumlnermeyi okumak iccedilin ruloyu ne kadar accedil-ması gerektiğini bilirdi [ p n ]

Oumlklidrsquoin yazdıklarının ccedileşitli suumlzgeccedillerden geccedilmiş ha-line ulaşabiliyoruz Oumlğelerin M Ouml civarında yazılmışolması gerekir Bizim kullandığımız rsquote yayınlananHeiberg versiyonu onuncu yuumlzyılda Vatikanrsquoda yazılan birelyazmasına dayanmaktadır

Analiz

Oumlğeler in her oumlnermesi bir problem veya bir teoremolarak anlaşılabilir MS civarında yazan İskenderi-yeli Pappus bu ayrımı tarif ediyor [ pp ndash]

Geometrik araştırmada daha teknik terimleri ter-cih edenler

bull problem (πρόβλημα) terimini iccedilinde [birşey]yapılması veya inşa edilmesi oumlnerilen [bir ouml-nerme] anlamında ve

bull teorem (θεώρημα) terimini iccedilinde belirli birhipotezin sonuccedillarının ve gerekliliklerinin in-celendiği [bir oumlnerme] anlamında

kullanırlar ama antiklerin bazıları bunlarıntuumlmuumlnuuml problem bazıları da teorem olarak tarifetmiştir

Kısaca bir problem birşey yapmayı oumlnerir bir teo-rem birşeyi goumlrmeyi (Yunancada Teorem kelimesi dahagenel olarak lsquobakılmış olanrsquo anlamındadır ve θεάομαι lsquobakrsquofiilyle ilgilidir burdan ayrıca θέατρον lsquotheaterrsquo kelimesi detuumlremiştir)

İster bir problem ister bir teorem olsun bir oumlnermemdashya da daha tam anlamıyla bir oumlnermenin metni mdashaltıparccedilaya kadar ayrılıp analiz edilebilir Oumlklidrsquoin Heath ccedile-virisinin The Green Lion baskısı [ p xxiii] bu analiziProclusrsquoun Commentary on the First Book of Euclidrsquos El-ements [ p ] kitabında bulunan haliyle tarif ederMS beşinci yuumlzyılda Proclus şoumlyle yazmıştır

Buumltuumln parccedilalarıyla donatılmış her problem ve teo-rem aşağıdaki oumlğeleri iccedilermelidir

) bir ilan (πρότασις)

) bir accedilıklama (ἔκθεσις)) bir belirtme (διορισμός)) bir hazırlama (κατασκευή)) bir goumlsteri (ἀπόδειξις) and) bir bitirme (συμπέρασμα)

Bunlardan ilan verileni ve bundan ne sonuccedil eldeedileceğini belirtir ccediluumlnkuuml muumlkemmel bir ilan buiki parccedilanın ikisini de iccedilerir Accedilıklama verileniayrıca ele alır ve bunu daha sonra incelemede kul-lanılmak uumlzere hazırlar Belirtme elde edileceksonucu ele alır ve onun ne olduğunu kesin bir şek-ilde accedilıklar Hazırlama elde edilecek sonuca ulaş-mak iccedilin verilende neyin eksik olduğunu soumlylerGoumlsteri oumlnerilen ccedilıkarımı kabul edilen oumlnermeler-den bilimsel akıl yuumlruumltmeyle oluşturur Bitirmeilana geri doumlnerek ispatlanmış olanı onaylar

Bir problem veya teoremin parccedilaları arasında enoumlnemli olanları her zaman bulunan ilan goumlsterive bitirmedir

Biz de Proclusrsquoun analizini aşağıdaki anlamıyla kul-lanacağız

İlan bir oumlnermenin harfli diagrama goumlnderme yap-mayan genel beyanıdır Bu beyan bir doğru veya uumlccedilgengibi bir nesne hakkındadır

Accedilıklamada bu nesne diagramla harfler aracılığıylaoumlzdeşleştirilir Bu nesnenin varlığı uumlccediluumlncuuml tekil emirkipinde bir fiil ile oluşturulur

(a) Belirtme bir problemde nesne ile ilgili neyapılacağını soumlyler ve δεῖ δὴ kelimeleriyle başlar Buradaδεῖ lsquogereklidirrsquo δή ise lsquoşimdirsquo anlamındadır

Proclus Bizans (yani Konstantinapolis şimdi İstanbul) doğum-ludur ama aslında Likyalıdır ve ilk eğitimini Ksantosrsquota almıştır

Felsefe oumlğrenmek iccedilin İskenderiyersquoye ve sonra da Atinarsquoya gitmiştir[ p xxxix]

(b) Bir teoremde belirtme nesneyle ilgili neyin ispat-lanacağını soumlyler ve lsquoİddia ediyorum kirsquo anlamına gelenλέγω ὅτι kelimeleriyle başlar Aynı ifade bir problemdede belirtmeye ek olarak goumlsterinin başında hazırlamanınsonunda goumlruumllebilir

Hazırlamada eğer varsa ikinci kelime γάρ onay-layıcı bir zarf ve sebep belirten bir bağlaccediltır Bu kelimeyicuumlmlenin birinci kelimesi lsquoccediluumlnkuumlrsquo olarak ccedileviriyoruz

Goumlsteri genellikle ἐπεί lsquoccediluumlnkuuml olduğundanrsquo ilgeciyle

başlar

Bitirme ilanı tekrarlar ve genellikle lsquodolayısıylarsquo il-gecini iccedilerir Tekrarlanan ilandan sonra bitirme aşağıdakiiki kalıptan biriyle sonlanır

(a) ὅπερ ἔδει ποιῆσαι lsquoyapılması gereken tam buydursquo(problemlerde)

(b) ὅπερ ἔδει δεῖξαι lsquogoumlsterilmesi gereken tam buydursquo (teo-remlerde) Latince quod erat demonstrandum veya qed

Dil

Oumlklidrsquoin kullandığı dil Antik Yunancadır Bu dil Hint-Avrupa dilleri ailesindendir İngilizce de bu ailedendir an-cak Tuumlrkccedile değildir Fakat bazı youmlnlerden Tuumlrkccedile Yunan-

caya İngilizceden daha yakındır İngilizce ve Tuumlrkccedileninguumlnuumlmuumlz bilimsel terminolojisinin koumlkleri genellikle Yu-nancadır

buumlyuumlk kuumlccediluumlk okunuş isimΑ α a alfaΒ β b betaΓ γ g gammaΔ δ d deltaΕ ε e epsilonΖ ζ z (ds) zetaΗ η ecirc (uzun e) etaΘ θ th thetaΙ ι i iota (yota)Κ κ k kappaΛ λ l lambdaΜ μ m muumlΝ ν n nuumlΞ ξ ks ksiΟ ο o (kısa) omikronΠ π p piΡ ρ r rho (ro)Σ σv ς s sigmaΤ τ t tauΥ υ y uuml uumlpsilonΦ φ f phiΧ χ h (kh) khiΨ ψ ps psiΩ ω ocirc (uzun o) omega

Table Yunan alfabesi

Chapter

Elements

lsquoDefinitionsrsquo

Boundaries ῞Οροι Sınırlar

[] A point is Σημεῖόν ἐστιν Bir nokta[that] whose part is nothing οὗ μέρος οὐθέν parccedilası hiccedilbir şey olandır

[] A line Γραμμὴ δὲ Bir ccedilizgilength without breadth μῆκος ἀπλατές ensiz uzunluktur

[] Of a line Γραμμῆς δὲ Bir ccedilizgininthe extremities are points πέρατα σημεῖα uccedillarındakiler noktalardır

[] A straight line is Εὐθεῖα γραμμή ἐστιν Bir doğruwhatever [line] evenly ἥτις ἐξ ἴσου uumlzerindeki noktalara hizalı uzanan birwith the points of itself τοῖς ἐφ᾿ ἑαυτῆς σημείοις ccedilizgidirlies κεῖται

[] A surface is ᾿Επιφάνεια δέ ἐστιν Bir yuumlzeywhat has length and breadth only ὃ μῆκος καὶ πλάτος μόνον ἔχει sadece eni ve boyu olandır

[] Of a surface ᾿Επιφανείας δὲ Bir yuumlzeyinthe boundaries are lines πέρατα γραμμαί uccedillarındakiler ccedilizgilerdir

[] A plane surface is ᾿Επίπεδος ἐπιφάνειά ἐστιν Bir duumlzlemwhat [surface] evenly ἥτις ἐξ ἴσου uumlzerindeki doğruların noktalarıylawith the points of itself ταῖς ἐφ᾿ ἑαυτῆς εὐθείαις hizalı uzanan bir yuumlzeydirlies κεῖται

[] A plane angle is ᾿Επίπεδος δὲ γωνία ἐστὶν Bir duumlzlem accedilısı ἡin a plane ἐν ἐπιπέδῳ bir duumlzlemdetwo lines taking hold of one another δύο γραμμῶν ἁπτομένων ἀλλήλων kesişen ve aynı doğru uumlzerinde uzan-and not lying on a straight καὶ μὴ ἐπ᾿ εὐθείας κειμένων mayanto one another πρὸς ἀλλήλας iki ccedilizginin birbirine goumlre eğikliğidirthe inclination of the lines τῶν γραμμῶν κλίσις

[] Whenever the lines containing the ῞Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν Ve accedilıyı iccedileren ccedilizgilerangle γραμμαὶ birer doğru olduğu zaman

be straight εὐθεῖαι ὦσιν duumlzkenar denir accedilıyarectilineal is called the angle εὐθύγραμμος καλεῖται ἡ γωνία

[] Whenever ῞Οταν δὲ Bir doğrua straight εὐθεῖα başka bir doğrunun uumlzerine yerleşip

The usual translation is lsquodefinitionsrsquo but what follow are notreally definitions in the modern sense

Presumably subject and predicate are inverted here so the sense

is that of lsquoA point is that of which nothing is a partrsquoThere is no way to put lsquothersquo here to parallel the Greek

standing on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα birbirine eşit bitişik accedilılar oluştur-the adjacent angles τὰς ἐφεξῆς γωνίας duğundaequal to one another make ἴσας ἀλλήλαις ποιῇ eşit accedilıların her birine dik accedilıright ὀρθὴ ve diğerinin uumlzerinde duran doğruyaeither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστι daand καὶ uumlzerinde durduğu doğruya bir dikthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖα doğru deniris called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

[] An obtuse angle is Αμβλεῖα γωνία ἐστὶν Bir geniş accedilıthat [which is] greater than a right ἡ μείζων ὀρθῆς buumlyuumlk olandır bir dik accedilıdan

[] Acute ᾿Οξεῖα δὲ Bir dar accedilıthat less than a right ἡ ἐλάσσων ὀρθῆς kuumlccediluumlk olandır bir dik accedilıdan

[] A boundary is ῞Ορος ἐστίν ὅ τινός ἐστι πέρας Bir sınırwhis is a limit of something bir şeyin ucunda olandır

[] A figure is Σχῆμά ἐστι Bir figuumlrwhat is contained by some boundary τὸ ὑπό τινος ἤ τινων ὅρων πε- bir sınır tarafından veya sınırlarca

or boundaries ριεχόμενον iccedilerilendir

[] A circle is Κύκλος ἐστὶ Bir dairea plane figure σχῆμα ἐπίπεδον duumlzlemdekicontained by one line ὑπὸ μιᾶς γραμμῆς περιεχόμενον bir ccedilizgice iccedilerilen[which is called the circumference] [ἣ καλεῖται περιφέρεια] [bu ccedilizgiye ccedilember denir]to which πρὸς ἣν bir figuumlrduumlr oumlyle kifrom one point ἀφ᾿ ἑνὸς σημείου figuumlruumln iccedilerisindekiof those lying inside of the figure τῶν ἐντὸς τοῦ σχήματος κειμένων noktaların birindenall straights falling πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι ccedilizgi uumlzerine gelen[to the circumference of the circle] [πρὸς τὴν τοῦ κύκλου περιφέρειαν] tuumlm doğrularare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittir

[] A center of the circle Κέντρον δὲ τοῦ κύκλου Ve o noktaya dairenin merkezi denirthe point is called τὸ σημεῖον καλεῖται

[] A diameter of the circle is Διάμετρος δὲ τοῦ κύκλου ἐστὶν Bir dairenin bir ccedilapısome straight εὐθεῖά τις dairenin merkezinden geccedilipdrawn through the center διὰ τοῦ κέντρου ἠγμένη her iki tarafta daand bounded καὶ περατουμένη dairenin ccedilevresindeki ccedilemberceto either parts εφ᾿ ἑκάτερα τὰ μέρη sınırlananby the circumference of the circle ὑπὸ τῆς τοῦ κύκλου περιφερείας bir doğrudurwhich also bisects the circle ἥτις καὶ δίχα τέμνει τὸν κύκλον ve boumlyle bir doğru daireyi ikiye boumller

[] A semicircle is ῾Ημικύκλιον δέ ἐστι Bir yarıdairethe figure contained τὸ περιεχόμενον σχῆμα bir ccedilapby the diameter ὑπό τε τῆς διαμέτρου ve onun kestiği bir ccedilevreceand the circumference taken off by it καὶ τῆς ἀπολαμβανομένης ὑπ᾿ αὐτῆς πε- iccedilerilen figuumlrduumlr ve yarıdaireninA center of the semicircle [is] the same ριφερείας merkezi o dairenin merkeziylewhich is also of the circle κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό aynıdır

ὃ καὶ τοῦ κύκλου ἐστίν

[] Rectilineal figures are Σχήματα εὐθύγραμμά ἐστι Duumlzkenar figuumlr lerthose contained by straights τὰ ὑπὸ εὐθειῶν περιεχόμενα doğrularca iccedilerilenlerdir Uumlccedilkenartriangles by three τρίπλευρα μὲν τὰ ὑπὸ τριῶν figuumlrler uumlccedil doumlrtkenar figuumlr-quadrilaterals by four τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων ler doumlrt ve ccedilokkenar figuumlrlerpolygons by more than four πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσ- ise doumlrtten daha fazla doğrucastraights contained σάρων iccedilerilenlerdir

This definition is quoted in Proposition In Greek what is repeated is not lsquoboundaryrsquo but lsquosomersquoNone of the terms defined in this section is preceeded by a defi-

nite article In particular what is being defined here is not the center

of a circle but a center However it is easy to show that the centerof a given circle is unique also in Proposition III Euclid finds thecenter of a given circle

CHAPTER ELEMENTS

εὐθειῶν περιεχόμενα

[] There being trilateral figures ῶν δὲ τριπλεύρων σχημάτων Uumlccedilkenar figuumlrlerdenan equilateral triangle is ἰσόπλευρον μὲν τρίγωνόν ἐστι bir eşkenar uumlccedilgenthat having three sides equal τὸ τὰς τρεῖς ἴσας ἔχον πλευράς uumlccedil kenarı eşit olanisosceles having only two sides equal ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔ- ikizkenar eşit iki kenarı olanscalene having three unequal sides χον πλευράς ccedileşitkenar uumlccedil kenarı eşit olmayandır

σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχονπλευράς

[] Yet of trilateral figures ῎Ετι δὲ τῶν τριπλεύρων σχημάτων Ayrıca uumlccedilkenar figuumlrlerdena right-angled triangle is ὀρθογώνιον μὲν τρίγωνόν ἐστι bir dik uumlccedilgenthat having a right angle τὸ ἔχον ὀρθὴν γωνίαν bir dik accedilısı olanobtuse-angled having an obtuse an- ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γω- geniş accedilılı bir geniş accedilısı olan

gle νίαν dar accedilılı uumlccedil accedilısı dar accedilı olandıracute-angled having three acute an- ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον

gles γωνίας

[] Of quadrilateral figures Τὼν δὲ τετραπλεύρων σχημάτων Doumlrtkenar figuumlrlerdena square is τετράγωνον μέν ἐστιν bir karewhat is equilateral and right-angled ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον hem eşit kenar hem de dik-accedilılı olanan oblong ἑτερόμηκες δέ bir dikdoumlrtgenright-angled but not equilateral ὃ ὀρθογώνιον μέν οὐκ ἰσόπλευρον δέ dik-accedilılı olan ama eşit kenar olmayana rhombus ῥόμβος δέ bir eşkenar doumlrtgenequilateral ὃ ἰσόπλευρον μέν eşit kenar olanbut not right-angled οὐκ ὀρθογώνιον δέ ama dik-accedilılı olmayanrhomboid ῥομβοειδὲς δὲ bir paralelkenarhaving opposite sides and angles τὸ τὰς ἀπεναντίον πλευράς τε καὶ γω- karşılıklı kenar ve accedilıları eşit olan

equal νίας ἴσας ἀλλήλαις ἔχον ama eşit kenar ve dik-accedilılı olmayandırwhich is neither equilateral nor right- ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώ- Ve bunların dışında kalan doumlrtke-

angled νιον narlara yamuk denilsinand let quadrilaterals other than these τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια

be called trapezia καλείσθω

[] Parallels are Παράλληλοί εἰσιν Paralellerstraights whichever εὐθεῖαι αἵτινες aynı duumlzlemde bulunanbeing in the same plane ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι ve her iki youmlnde deand extended to infinity καὶ ἐκβαλλόμεναι εἰς ἄπειρον sınırsızca uzatıldıklarındato either parts ἐφ᾿ ἑκάτερα τὰ μέρη hiccedilbir noktada kesişmeyento neither [parts] fall together with ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις doğrulardır

one another

As in Turkish so in Greek a plural subject can take a singularverb when the subject is of the neuter gender in Greek or namesinanimate objects in Turkish

To maintain the parallelism of the Greek we could (like Heath)use lsquotrilateralrsquo lsquoquadrilateralrsquo and lsquomultilateralrsquo instead of lsquotrianglersquolsquoquadrilateralrsquo and lsquopolygonrsquo Today triangles and quadrilateralsare polygons For Euclid they are not you never call a triangle apolygon because you can give the more precise information that itis a triangle

Postulates

Postulates Αἰτήματα Postulatlar

Let it have been postulated ᾿Ηιτήσθω Postulat olarak kabul edilsinfrom any point ἀπὸ παντὸς σημείου herhangi bir noktadanto any point ἐπὶ πᾶν σημεῖον herhangi bir noktayaa straight line εὐθεῖαν γραμμὴν bir doğruto draw ἀγαγεῖν ccedilizilmesi

Also a bounded straight Καὶ πεπερασμένην εὐθεῖαν Ve sonlu bir doğrununcontinuously κατὰ τὸ συνεχὲς kesiksiz şekildein a straight ἐπ᾿ εὐθείας bir doğrudato extend ἐκβαλεῖν uzatılması

Also to any center Καὶ παντὶ κέντρῳ Ve her merkezand distance καὶ διαστήματι ve uzunluğaa circle κύκλον bir daireto draw γράφεσθαι ccedilizilmesi

Also all right angles Καὶ πάσας τὰς ὀρθὰς γωνίας Ve buumltuumln dik accedilılarınequal to one another ἴσας ἀλλήλαις bir birine eşitto be εἶναι olduğu

Also if in two straight lines Καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα Ve iki doğruyufalling ἐμπίπτουσα kesen bir doğrununthe interior angles to the same parts τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας aynı tarafta oluşturduğuless than two rights make δύο ὀρθῶν ἐλάσσονας ποιῇ iccedil accedilılar iki dik accedilıdan kuumlccediluumlksethe two straights extended ἐκβαλλομένας τὰς δύο εὐθείας bu iki doğrununto infinity ἐπ᾿ ἄπειρον sınırsızca uzatıldıklarındafall together συμπίπτειν accedilılarınto which parts are ἐφ᾿ ἃ μέρη εἰσὶν iki dik accedilıdan kuumlccediluumlk olduğu taraftathe less than two rights αἱ τῶν δύο ὀρθῶν ἐλάσσονες kesişeceği

CHAPTER ELEMENTS

Common Notions

Common notions Κοιναὶ ἔννοιαι Genel Kavramlar

Equals to the same Τὰ τῷ αὐτῷ ἴσα Aynı şeye eşitleralso to one another are equal καὶ ἀλλήλοις ἐστὶν ἴσα birbirlerine de eşittir

Also if to equals Καὶ ἐὰν ἴσοις Eğer eşitlereequals be added ἴσα προστεθῇ eşitler eklenirsethe wholes are equal τὰ ὅλα ἐστὶν ἴσα elde edilenler de eşittir

Also if from equals αὶ ἐὰν ἀπὸ ἴσων Eğer eşitlerdenequals be taken away ἴσα ἀφαιρεθῇ eşitler ccedilıkartılırsathe remainders are equal τὰ καταλειπόμενά ἐστιν ἴσα kalanlar eşittir

Also things applying to one another Καὶ τὰ ἐφαρμόζοντα ἐπ᾿ ἀλλήλα Birbiriyle ccedilakışan şeylerare equal to one another ἴσα ἀλλήλοις ἐστίν birbirine eşittir

Also the whole Καὶ τὸ ὅλον Buumltuumlnthan the part is greater τοῦ μέρους μεῖζόν [ἐστιν] parccediladan buumlyuumlktuumlr

On ᾿Επὶ Verilmiş sınırlanmış doğruyathe given bounded straight τῆς δοθείσης εὐθείας πεπερασμένης eşkenar uumlccedilgenfor an equilateral triangle τρίγωνον ἰσόπλευρον inşa edilmesito be constructed συστήσασθαι

Let be ῎Εστω Verilmişthe given bounded straight ἡ δοθεῖσα εὐθεῖα πεπερασμένη sınırlanmış doğruΑΒ ἡ ΑΒ ΑΒ olsun

It is necessary then Δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἐπὶ τῆς ΑΒ εὐθείας ΑΒ doğrusunafor an equilateral triangle τρίγωνον ἰσόπλευρον eşkenar uumlccedilgeninto be constructed συστήσασθαι inşa edilmesi

To center Α Κέντρῳ μὲν τῷ Α Α merkezineat distance ΑΒ διαστήματι δὲ τῷ ΑΒ ΑΒ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΒΓΔ ὁ ΒΓΔ ΒΓΔand moreover καὶ πάλιν ve yineto center Β κέντρῳ μὲν τῷ Β Β merkezineat distance ΒΑ διαστήματι δὲ τῷ ΒΑ ΒΑ uzaklığında olansuppose a circle has been drawn κύκλος γεγράφθω ccedilember ccedilizilmiş olsun[namely] ΑΓΕ ὁ ΑΓΕ ΑΓΕand from the point Γ καὶ ἀπὸ τοῦ Γ σημείου ccedilemberlerin kesiştiğiwhere the circles cut one another καθ᾿ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι Γ noktasındanto the points Α and Β ἐπί τὰ Α Β σημεῖα Α Β noktalarınasuppose there have been joined ἐπεζεύχθωσαν ΓΑ ΓΒ doğruları birleştirilmiş olsunthe straights ΓΑ and ΓΒ εὐθεῖαι αἱ ΓΑ ΓΒ

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΓΔΒ κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου ΓΔΒ ccedilemberinin merkezi olduğu iccedilinequal is ΑΓ to ΑΒ ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ΑΓ ΑΒ doğrusuna eşittirmoreover πάλιν Dahasısince the point Β ἐπεὶ τὸ Β σημεῖον Β noktası ΓΑΕ ccedilemberinin merkeziis the center of the circle ΓΑΕ κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου olduğu iccedilinequal is ΒΓ to ΒΑ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ ΒΓ ΒΑ doğrusuna eşittir

Heathrsquos translation has the indefinite article lsquoarsquo here in ac-cordance with modern mathematical practice However Eucliddoes use the Greek definite article here just as in the exposition(see sect) In particular he uses the definite article as a genericarticle which lsquomakes a single object the representative of the entireclassrsquo [ para p ] English too has a generic use of thedefinite article lsquoto indicate the class or kind of objects as in thewell-known aphorism The child is the father of the manrsquo [p ] (However the enormous Cambridge Grammar does notdiscuss the generic article in the obvious place [ pp ndash] By the way the lsquowell-known aphorismrsquo is by Wordsworthsee httpenwikisourceorgwikiOde_Intimations_of_

Immortality_from_Recollections_of_Early_Childhood [accessedJuly ]) See note to Proposition below

The Greek form of the enunciation here is an infinitive clauseand the subject of such a clause is generally in the accusative case[ para p ] In English an infinitive clause with expressedsubject (as here) is always preceded by lsquoforrsquo [ p ]Normally such a clause in Greek or English does not stand by itselfas a complete sentence here evidently it is expected to Note thatthe Greek infinitive is thought to be originally a noun in the dativecase [ para p ] the English infinitive with lsquotorsquo would seemto be formed similarly

We follow Euclid in putting the verb (a third-person imperative)first but a smoother translation of the exposition here would be lsquoLetthe given finite straight line be ΑΒrsquo Heathrsquos version is lsquoLet AB bethe given finite straight linersquo By the argument of Netz [ pp ndash]this would appear to be a misleading translation if not a mistransla-tion Euclidrsquos expression ἡ ΑΒ lsquothe ΑΒrsquo must be understood as anabbreviation of ἡ εὐθεῖα γραμμὴ ἡ ΑΒ or ἡ ΑΒ εὐθεῖα γραμμή lsquothe

straight line ΑΒrsquo In Proposition XIII Euclid says ῎Εστω εὐθεῖα ἡΑΒ which Heath translates as lsquoLet AB be a straight linersquo but thenthis suggests the expansion lsquoLet the straight line AB be a straightlinersquo which does not make much sense Netzrsquos translation is lsquoLetthere be a straight line [namely] ABrsquo The argument is that Eucliddoes not use words to establish a correlation between letters like A

and B and points The correlation has already been established inthe diagram that is before us By saying ῎Εστω εὐθεῖα ἡ ΑΒ Euclidis simply calling our attention to a part of the diagram Now inthe present proposition Heathrsquos translation of the exposition is ex-panded to lsquoLet the straight line AB be the given finite straight linersquowhich does seem to make sense at least if it can be expanded furtherto lsquoLet the finite straight line AB be the given finite straight linersquoBut unlike AB the given finite straight line was already mentionedin the enunciation so it is less misleading to name this first in theexposition

Slightly less literally lsquoIt is necessary that on the straight ΑΒan equilateral triangle be constructedrsquo

Instead of lsquosuppose there have been joinedrsquo we could write lsquoletthere have been joinedrsquo However each of these translations of aGreek third-person imperative begins with a second-person imper-ative (because there is no third-person imperative form in Englishexcept in some fixed forms like lsquoGod bless yoursquo) The logical sub-ject of the verb lsquohave been joinedrsquo is lsquothe straight ΑΒrsquo since thiscomes after the verb it would appear to be an extraposed subject inthe sense of the Cambridge Grammar of the English Language [ p ] Then the grammatical subject of lsquohave been joinedrsquo islsquotherersquo used as a dummy but it will not always be appropriate touse a dummy in such situations [ p ndash]

CHAPTER ELEMENTS

And ΓΑ was shown equal to ΑΒ ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση Ve ΓΑ doğrusunun ΑΒ doğrusuna eşittherefore either of ΓΑ and ΓΒ to ΑΒ ἑκατέρα ἄρα τῶν ΓΑ ΓΒ τῇ ΑΒ olduğu goumlsterilmiştiis equal ἐστιν ἴση O zaman ΓΑ ΓΒ doğrularının her biriBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα ΑΒ doğrusuna eşittirare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlartherefore also ΓΑ is equal to ΓΒ καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστιν ἴση birbirine eşittirTherefore the three ΓΑ ΑΒ and ΒΓ αἱ τρεῖς ἄρα αἱ ΓΑ ΑΒ ΒΓ O zaman ΓΑ ΓΒ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν O zaman o uumlccedil doğru ΓΑ ΑΒ ΒΓ

birbirine eşittir

Equilateral therefore ᾿Ισόπλευρον ἄρα Eşkenardır dolayısıylais triangle ΑΒΓ ἐστὶ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeniAlso it has been constructed καὶ συνέσταται ve inşa edilmiştiron the given bounded straight ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης verilmiş sınırlanmışΑΒ τῆς ΑΒ6 ΑΒ doğrusunamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ Ε

At the given point Πρὸς τῷ δοθέντι σημείῳ Verilmiş noktayaequal to the given straight τῇ δοθείσῃ εὐθείᾳ ἴσην verilmiş doğruya eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

Let be ῎Εστω Verilmiş nokta Α olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilmiş doğru ΒΓand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ

It is necessary then δεῖ δὴ Gereklidirat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the given straight ΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην ΒΓ doğrusuna eşit olanfor a straight to be placed εὐθεῖαν θέσθαι bir doğrunun konulması

For suppose there has been joined ᾿Επεζεύχθω γὰρ Ccediluumlnkuuml birleştirilmiş olsunfrom the point Α to the point Β ἀπὸ τοῦ Α σημείου ἐπί τὸ Β σημεῖον Α noktasından Β noktasınaa straight ΑΒ εὐθεῖα ἡ ΑΒ ΑΒ doğrusuand there has been constructed on it καὶ συνεστάτω ἐπ᾿ αὐτῆς ve bu doğru uumlzerine inşa edilmiş olsunan equilateral triangle ΔΑΒ τρίγωνον ἰσόπλευρον τὸ ΔΑΒ eşkenar uumlccedilgen ΔΑΒand suppose there have been extended καὶ ἐκβεβλήσθωσαν ve uzatılmış olsunon a straight with ΔΑ and ΔΒ ἐπ᾿ εὐθείας ταῖς ΔΑ ΔΒ ΔΑ ΔΒ doğrularındanthe straights ΑΕ and ΒΖ εὐθεῖαι αἱ ΑΕ ΒΖ ΑΕ ΒΖ doğrularıand to the center Β καὶ κέντρῳ μὲν τῷ Β ve Β merkezineat distance ΒΓ διαστήματι δὲ τῷ ΒΓ ΒΓ uzaklığındasuppose a circle has been drawn κύκλος γεγράφθω ccedilizilmiş olsunΓΗΘ ὁ ΓΗΘ ΓΗΘ ccedilemberi ve yine Δ merkezineand again to the center Δ καὶ πάλιν κέντρῳ τῷ Δ ΔΗ uzaklığındaat distance ΔΗ καὶ διαστήματι τῷ ΔΗ ccedilizilmiş olsunsuppose a circle has been drawn κύκλος γεγράφθω ΗΚΛ ccedilemberi

Normally Heiberg puts a semicolon at this position Perhapshe has a period here only because he has bracketed the followingwords (omitted here) lsquoTherefore on a given bounded straight

an equilateral triangle has been constructedrsquo According to Heibergthese words are found not in the manuscripts of Euclid but inProclusrsquos commentary [ p ] alone

ΗΚΛ ὁ ΗΚΛ

Since then the point Β is the center ᾿Επεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ Β noktası ΓΗΘ ccedilemberinin merkeziof ΓΗΘ τοῦ ΓΗΘ olduğu iccedilinΒΓ is equal to ΒΗ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ ΒΓ ΒΗ doğrusuna eşittirMoreover πάλιν Yinesince the point Δ is the center ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ Δ noktası ΗΚΛ ccedilemberinin merkeziof the circle ΚΗΛ τοῦ ΗΚΛ κύκλου olduğu iccedilinequal is ΔΛ to ΔΗ ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ ΔΛ ΔΗ doğrusuna eşittirof these the [part] ΔΑ to ΔΒ ὧν ἡ ΔΑ τῇ ΔΒ ve (birincinin) ΔΑ parccedilasıis equal ἴση ἐστίν (ikincinin) ΔΒ parccedilasına eşittirTherefore the remainder ΑΛ λοιπὴ ἄρα ἡ ΑΛ Dolayısıyla ΑΛ kalanıto the remainder ΒΗ λοιπῇ τῇ ΒΗ ΒΗ kalanınais equal ἐστιν ἴση eşittirBut ΒΓ was shown equal to ΒΗ ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση Ve ΒΓ doğrusunun ΒΗ doğrusunaTherefore either of ΑΛ and ΒΓ to ΒΗ ἑκατέρα ἄρα τῶν ΑΛ ΒΓ τῇ ΒΗ eşit olduğu goumlsterilmiştiis equal ἐστιν ἴση Dolayısıyla ΑΛ ΒΓ doğrularının herBut equals to the same τὰ δὲ τῷ αὐτῷ ἴσα biri ΒΗ doğrusuna eşittiralso are equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα Ama aynı şeye eşit olanlar birbirineAnd therefore ΑΛ is equal to ΒΓ καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστιν ἴση eşittir

Ve dolayısıyla ΑΛ da ΒΓ doğrusunaeşittir

Therefore at the given point Α Πρὸς ἄρα τῷ δοθέντι σημείῳ Dolayısıyla verilmiş Α noktasınaequal to the given straight ΒΓ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση verilmiş ΒΓ doğrusuna eşit olanthe straight ΑΛ is laid down εὐθεῖα κεῖται ἡ ΑΛ ΑΛ doğrusu konulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε Ζ

Η

Θ

Κ

Λ

Two unequal straights being given Δύο δοθεισῶν εὐθειῶν ἀνίσων İki eşit olmayan doğru verilmiş isefrom the greater ἀπὸ τῆς μείζονος daha buumlyuumlktenequal to the less τῇ ἐλάσσονι ἴσην daha kuumlccediluumlğe eşit olana straight to take away εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let be ῎Εστωσαν İki verilmiş doğruthe two given unequal straights αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι ΑΒ ΓΑΒ and Γ αἱ ΑΒ Γ olsunlarof which let the greater be ΑΒ ὧν μείζων ἔστω ἡ ΑΒ daha buumlyuumlğuuml ΑΒ olsun

It is necessary then δεῖ δὴ Gereklidirfrom the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundan

The phrase ἐπ᾿ εὐθείας will recur a number of times The ad-jective which is feminine here appears to be a genitive singularthough it could be accusative plural

Since Γ is given the feminine gender in the Greek this is a signthat Γ is indeed a line and not a point See the Introduction

CHAPTER ELEMENTS

equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴσην daha kuumlccediluumlk olan Γ doğrusuna eşit olanto take away a straight εὐθεῖαν ἀφελεῖν bir doğru kesmek

Let there be laid down Κείσθω Konulsunat the point Α πρὸς τῷ Α σημείῳ Α noktasınaequal to the line Γ τῇ Γ εὐθείᾳ ἴση Γ doğrusuna eşit olanΑΔ ἡ ΑΔ ΑΔ doğrusuand to center Α καὶ κέντρῳ μὲν τῷ Α Ve Α merkezineat distance ΑΔ διαστήματι δὲ τῷ ΑΔ ΑΔ uzaklığında olansuppose circle ΔΕΖ has been drawn κύκλος γεγράφθω ὁ ΔΕΖ ΔΕΖ ccedilemberi ccedilizilmiş olsun

And since the point Α Καὶ ἐπεὶ τὸ Α σημεῖον Ve Α noktasıis the center of the circle ΔΕΖ κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου ΔΕΖ ccedilemberinin merkezi olduğu iccedilinequal is ΑΕ to ΑΔ ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ ΑΕ ΑΔ doğrusuna eşittirBut Γ to ΑΔ is equal ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση Ama Γ ΑΔ doğrusuna eşittirTherefore either of ΑΕ and Γ ἑκατέρα ἄρα τῶν ΑΕ Γ Dolayısıyla ΑΕ Γ doğrularının heris equal to ΑΔ τῇ ΑΔ ἐστιν ἴση biriand so ΑΕ is equal to Γ ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση ΑΔ doğrusuna eşittir

Sonuccedil olarakΑΕ Γ doğrusuna eşittir

Therefore two unequal straights Δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν Dolayısıyla iki eşit olmayan ΑΒ Γbeing given ΑΒ and Γ ΑΒ Γ doğrusu verilmiş ise

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ daha buumlyuumlk olan ΑΒ doğrusundanan equal to the less Γ τῇ ἐλάσσονι τῇ Γ ἴση daha kuumlccediluumlk olan Γ doğrusuna eşit olanhas been taken away [namely] ΑΕ ἀφῄρηται ἡ ΑΕ ΑΕ doğrusu kesilmiştimdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

ΓΔ

Ε

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgendetwo sides τὰς δύο πλευρὰς iki kenarto two sides [ταῖς] δυσὶ πλευραῖς iki kenarahave equal ἴσας ἔχῃ eşit olursaeither [side] to either ἑκατέραν ἑκατέρᾳ (her biri birine)and angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ ve accedilı accedilıya eşit olursamdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν (yani eşit doğrular tarafından

straights περιεχομένην iccedilerilen)contained καὶ τὴν βάσιν τῂ βάσει hem taban tabanaalso base to base ἴσην ἕξει eşit olacakthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ hem uumlccedilgen uumlccedilgeneand the triangle to the triangle ἴσον ἔσται eşit olacakwill be equal καὶ αἱ λοιπαὶ γωνίαι hem de geriye kalan accedilılarand the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarato the remaining angles ἴσαι ἔσονται eşit olacakwill be equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν (yani) eşit kenarları goumlrenler

mdashthose that the equal sides subtend

Let be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ (adlarında) iki uumlccedilgenthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓto the two sides ΔΕ and ΔΖ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ ΔΖ ΔΕ ΔΖ iki kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ and ΑΓ to ΔΖ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ (şoumlyle ki) ΑΒΔΕ kenarına ve ΑΓΔΖand angle ΒΑΓ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ kenarınato ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ ve ΒΑΓ (tarafından iccedilerilen) accedilısıequal ἴσην ΕΔΖ accedilısına

eşit olan

I say that λέγω ὅτι İddia ediyorum kithe base ΒΓ is equal to the base ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν ΒΓ tabanı eşittir ΕΖ tabanınaand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgeniwill be equal to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται eşit olacak ΔΕΖ uumlccedilgenineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacak geriyeto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (şoumlyle ki) eşit kenarları goumlrenlerthose that equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΒΓ ΔΕΖ accedilısına[namely] ΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΓΒ ΔΖΕ accedilısınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgenito triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ΔΕΖ uumlccedilgenininand there being placed καὶ τιθεμένου ve yerleştirilirsethe point Α on the point Δ τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον Α noktası Δ noktasınaand the straight ΑΒ on ΔΕ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ ve ΑΒ doğrusu ΔΕ doğrusunaalso the point Β will apply to Ε ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε o zaman Β noktası yerleşecek Ε nok-by the equality of ΑΒ to ΔΕ διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ tasınaThen ΑΒ applying to ΔΕ ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ΑΒ doğrusunun ΔΕ doğrusuna eşitliğialso straight ΑΓ will apply to ΔΖ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ sayesindeby the equality διὰ τὸ ἴσην εἶναι Boumlylece ΑΒ doğrusunu yerleştirilinceof angle ΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ ΔΕ doğrusunaHence the point Γ to the point Ζ ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ΑΓ doğrusu uumlstuumlne gelecek ΔΖwill apply ἐφαρμόσει doğrusununby the equality again of ΑΓ to ΔΖ διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ ΒΑΓ accedilısının eşitliği sayesindeBut Β had applied to Ε ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει ΕΔΖ accedilısınaHence the base ΒΓ to the base ΕΖ ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ Dolayısıyla Γ noktası yerleşecek Ζwill apply ἐφαρμόσει noktasınaFor if εἰ γὰρ eşitliği sayesinde yine ΑΓ doğrusu-Β applying to Ε τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος nun ΔΖ doğrusunaand Γ to Ζ τοῦ δὲ Γ ἐπὶ τὸ Ζ Ama Β konuldu Ε noktasınathe base ΒΓ will not apply to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει Dolayısıyla ΒΓ tabanı uumlstuumlne gelecektwo straights will enclose a space δύο εὐθεῖαι χωρίον περιέξουσιν ΕΖ tabanınınwhich is impossible ὅπερ ἐστὶν ἀδύνατον Ccediluumlnkuuml eğer konulunca Β Ε nok-Therefore will apply ἐφαρμόσει ἄρα tasınabase ΒΓ to ΕΖ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ ve Γ Ζ noktasınaand will be equal to it καὶ ἴση αὐτῇ ἔσται ΒΓ tabanı yerleşmeyecekse ΕΖ ta-Hence triangle ΑΒΓ as a whole ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον banına

More smoothly lsquoIf two triangles have two sides equal to twosidesrsquo

That is lsquorespectivelyrsquo We could translate the Greek also aslsquoeach to eachrsquo but the Greek ἑκατέρος has the dual number asopposed to ἕκαστος lsquoeachrsquo The English form lsquoeitherrsquo is a remnantof the dual number

It appears that for Euclid things are never simply equal theyare equal to something Here the equal straights containing theangle are not equal to one another they are separately equal to thetwo straights in the other triangle

Here Euclidrsquos καί has a different meaning from the earlier in-stance now it shows the transition to the conclusion of the enunci-ation In fact the conclusion has the form καί καί καί Thisgeneral form might be translated as lsquoBoth and and rsquo Theword both properly refers to two things but the Oxford English Dic-tionary cites an example from Chaucer () where it refers to threethings lsquoBoth heaven and earth and searsquo The word both seems tohave entered English late from Old Norse it supplanted the earlierwordbo

CHAPTER ELEMENTS

to triangle ΔΕΖ as a whole ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον iki doğru ccedilevreleyecek bir alanwill apply ἐφαρμόσει imkansız olanand will be equal to it καὶ ἴσον αὐτῷ ἔσται Bu yuumlzden ΒΓ tabanı ccedilakışacak ΕΖand the remaining angles καὶ αἱ λοιπαὶ γωνίαι tabanıylato the remaining angles ἐπὶ τὰς λοιπὰς γωνίας ve eşit olacak onawill apply ἐφαρμόσουσι Dolayısıyla ΑΒΓ uumlccedilgeninin tamamıand be equal to them καὶ ἴσαι αὐταῖς ἔσονται uumlstuumlne gelecek ΔΕΖ uumlccedilgenininΑΒΓ to ΔΕΖ ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ tamamınaand ΑΓΒ to ΔΖΕ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ve eşit olacak ona

ve geriye kalan accedilılar uumlstuumlne geleceklergeriye kalan accedilıların

ve eşit olacaklar onlaraΑΒΓ ΔΕΖ accedilısınave ΑΓΒ ΔΖΕ accedilısına

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Dolayısıyla eğertwo sides τὰς δύο πλευρὰς iki uumlccedilgenin varsa iki kenarı eşit olanto two sides [ταῖς] δύο πλευραῖς iki kenarahave equal ἴσας ἔχῃ her bir (kenar) birineeither to either ἑκατέραν ἑκατέρᾳ ve varsa accedilıya eşit accedilısıand angle to angle have equal καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ (yani) eşit doğrularca iccedilerilenmdashthat which is by the equal τὴν ὑπὸ τῶν ἴσων εὐθειῶν hem tabana eşit tabanları olacak

straights περιεχομένην hem uumlccedilgen eşit olacak uumlccedilgenecontained καὶ τὴν βάσιν τῂ βάσει hem de geriye kalan accedilılar eşit olacakalso base to base ἴσην ἕξει geriye kalan accedilılarınthey will have equal καὶ τὸ τρίγωνον τῷ τριγώνῳ her biri birineand the triangle to the triangle ἴσον ἔσται (yani) eşit kenarları goumlrenlerwill be equal καὶ αἱ λοιπαὶ γωνίαι mdashgoumlsterilmesi gereken tam buyduand the remaining angles ταῖς λοιπαῖς γωνίαιςto the remaining angles ἴσαι ἔσονταιwill be equal ἑκατέρα ἑκατέρᾳeither to either ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσινmdashthose that the equal sides subtend ὅπερ ἔδει δεῖξαιmdashjust what it was necessary to show

Α

Β Γ

Δ

Ε Ζ

In isosceles triangles Τῶν ἰσοσκελῶν τριγώνων İkizkenar uumlccedilgenlerdethe angles at the base αἱ πρὸς τῇ βάσει γωνίαι tabandaki accedilılarare equal to one another ἴσαι ἀλλήλαις εἰσίν birbirine eşittirand καὶ vethe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν eşit doğrular uzatıldığındathe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι tabanın altında kalan accedilılarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται birbirine eşit olacaklar

Let there be ῎Εστω Verilmiş olsunan isosceles triangle ΑΒΓ τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ bir ΑΒΓ ikizkenar uumlccedilgenihaving equal ἴσην ἔχον ΑΒ kenarı eşit olan ΑΓ kenarınaside ΑΒ to side ΑΓ τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ ve varsayılsın ΒΔ ve ΓΕ doğrularınınand suppose have been extended καὶ προσεκβεβλήσθωσαν uzatılmış olduğu ΑΒ ve ΑΓon a straight with ΑΒ and ΑΓ ἐπ᾿ εὐθείας ταῖς ΑΒ ΑΓ doğrularından

Heath has coinciding here but the verb is just the active formof what in the passive is translated as being applied

More literally lsquoofrsquo

the straights ΒΔ and ΓΕ εὐθεῖαι αἱ ΒΔ ΓΕ

I say that λέγω ὅτι İddia ediyorum kiangle ΑΒΓ to angle ΑΓΒ ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ΑΒΓ accedilısı ΑΓΒ accedilısınais equal ἴση ἐστίν eşittirand ΓΒΔ to ΒΓΕ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ ve ΓΒΔ accedilısı eşittir ΒΓΕ accedilısına

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş ol-a random point Ζ on ΒΔ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ sunand there has been taken away καὶ ἀφῃρήσθω rastgele bir Ζ noktası ΒΔ uumlzerinndefrom the greater ΑΕ ἀπὸ τῆς μείζονος τῆς ΑΕ ve ΑΗto the less ΑΖ τῇ ἐλάσσονι τῇ ΑΖ buumlyuumlk olan ΑΕ doğrusundanan equal ΑΗ ἴση ἡ ΑΗ kuumlccediluumlk olan ΑΖ doğrusunun kesilmişiand suppose there have been joined καὶ ἐπεζεύχθωσαν olsunthe straights ΖΓ and ΗΒ αἱ ΖΓ ΗΒ εὐθεῖαι ve ΖΓ ile ΗΒ birleştirilmiş olsun

Since then ΑΖ is equal to ΑΗ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ Ccediluumlnkuuml o zaman ΑΖ eşittir ΑΗand ΑΒ to ΑΓ ἡ δὲ ΑΒ τῇ ΑΓ doğrusunaso the two ΑΖ and ΑΓ δύο δὴ αἱ ΖΑ ΑΓ ve ΑΒ doğrusu ΑΓ doğrusunato the two ΗΑ ΑΒ δυσὶ ταῖς ΗΑ ΑΒ boumlylece ΑΖ ve ΑΓ ikilisi eşit olacakwill be equal ἴσαι εἰσὶν ΗΑ ve ΑΒ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand they bound a common angle καὶ γωνίαν κοινὴν περιέχουσι ve sınırlandırırlar ortak bir accedilıyı[namely] ΖΑΗ τὴν ὑπὸ ΖΑΗ (yani) ΖΑΗ accedilısınıtherefore the base ΖΓ to the base ΗΒ βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ dolayısıyla ΖΓ tabanı eşittir ΗΒ ta-is equal ἴση ἐστίν banınaand triangle ΑΖΓ to triangle ΑΗΒ καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ve ΑΖΓ uumlccedilgeni eşit olacak ΑΗΒ uumlccedilge-will be equal ἴσον ἔσται nineand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve geriye kalan accedilılar eşit olacaklarto the remaining angles ταῖς λοιπαῖς γωνίαις geriye kalan accedilılarınwill be equal ἴσαι ἔσονται her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν ΑΓΖ accedilısı ΑΒΗ accedilısınaΑΓΖ to ΑΒΗ ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ ve ΑΖΓ accedilısı ΑΗΒ accedilısınaand ΑΖΓ to ΑΗΒ ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ Boumlylece ΑΖ buumltuumlnuumlnuumln eşitliği ΑΗAnd since ΑΖ as a whole καὶ ἐπεὶ ὅλη ἡ ΑΖ buumltuumlnuumlneto ΑΗ as a whole ὅλῃ τῇ ΑΗ ve bunların ΑΒ parccedilasının eşitliği ΑΓis equal ἐστιν ἴση parccedilasınaof which the [part] ΑΒ to ΑΓ is equal ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση gerektirir ΒΖ kalanının eşit olmasınıtherefore the remainder ΒΖ λοιπὴ ἄρα ἡ ΒΖ ΓΗ kalanınato the remainder ΓΗ λοιπῇ τῇ ΓΗ Ve ΖΓ doğrusunun goumlsterilmişti eşitis equal ἐστιν ἴση olduğu ΗΒ doğrusunaAnd ΖΓ was shown equal to ΗΒ ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση O zaman ΒΖ ve ΖΓ ikilisi eşittir ΓΗveThen the two ΒΖ and ΖΓ δύο δὴ αἱ ΒΖ ΖΓ ΗΒ ikilisininto the two ΓΗ and ΗΒ δυσὶ ταῖς ΓΗ ΗΒ her biri birineare equal ἴσαι εἰσὶν ve ΒΖΓ accedilısı ΓΗΒ accedilısınaeither to either ἑκατέρα ἑκατέρᾳ ve onların ortak tabanı ΒΓand angle ΒΖΓ καὶ γωνία ἡ ὑπὸ ΒΖΓ doğrusudurto angle ΓΗΒ γωνίᾳ τῃ ὑπὸ ΓΗΒ ve bu yuumlzden ΒΖΓ uumlccedilgeni eşit olacak[is] equal ἴση ΓΗΒ uumlccedilgenineand the common base of them is ΒΓ καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ ve geriye kalan accedilılar da eşit olacaklarand therefore triangle ΒΖΓ καὶ τὸ ΒΖΓ ἄρα τρίγωνον geriye kalan accedilılarınto triangle ΓΗΒ τῷ ΓΗΒ τριγώνῳ her biri birinewill be equal ἴσον ἔσται aynı kenarları goumlrenlerand the remaining angles καὶ αἱ λοιπαὶ γωνίαι Dolayısıyla ΖΒΓ eşittir ΗΓΒ accedilısınato the remaining angles ταῖς λοιπαῖς γωνίαις ve ΒΓΖ accedilısı ΓΒΗ accedilısınawill be equal ἴσαι ἔσονται Ccediluumlnkuuml goumlsterilmiş oldu ΑΒΗ accedilısınıneither to either ἑκατέρα ἑκατέρᾳ buumltuumlnuumlnuumln eşit olduğu ΑΓΖwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν accedilısının buumltuumlnuumlneEqual therefore is ἴση ἄρα ἐστὶν ve bunların ΓΒΗ parccedilasının (eşitliği)ΖΒΓ to ΗΓΒ ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ΒΓΖ parccedilasınaand ΒΓΖ to ΓΒΗ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ dolayısıyla ΑΒΓ kalanı eşittir ΑΓΒSince then angle ΑΒΗ as a whole ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία kalanına

CHAPTER ELEMENTS

to angle ΑΓΖ as a whole ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ve bunlar ΑΒΓ uumlccedilgeninin tabanıdırwas shown equal ἐδείχθη ἴση Ve ΖΒΓ accedilısının eşit olduğu goumlster-of which the [part] ΓΒΗ to ΒΓΖ ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ilmişti ΗΓΒ accedilısınais equal ἴση ve bunlar tabanın altındadırtherefore the remainder ΑΒΓ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓto the remainder ΑΓΒ λοιπῇ τῇ ὑπὸ ΑΓΒis equal ἐστιν ἴσηand they are at the base καί εἰσι πρὸς τῇ βάσειof the triangle ΑΒΓ τοῦ ΑΒΓ τριγώνουAnd was shown also ἐδείχθη δὲ καὶΖΒΓ equal to ΗΓΒ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴσηand they are under the base καί εἰσιν ὑπὸ τὴν βάσιν

Therefore in isosceles triangles Τῶν ἰσοσκελῶν τριγώνων Dolayısıyla bir ikizkenar uumlccedilgenin ta-the angles at the base αἱ πρὸς τῇ βάσει γωνίαι banındaki accedilılar birbirine eşit-are equal to one another ἴσαι ἀλλήλαις εἰσίν tirand καὶ ve eşit doğrular uzatıldığındathe equal straights being extended προσεκβληθεισῶν τῶν ἴσων εὐθειῶν tabanın altında kalan accedilılar birbirinethe angles under the base αἱ ὑπὸ τὴν βάσιν γωνίαι eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

Η

Ε

Ζ

If in a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν birbirine eşit iki accedilı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar da

angles πλευραὶ birbirine eşit olacaklarwill be equal to one another ἴσαι ἀλλήλαις ἔσονται

Let there be ῎Εστω Verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving equal ἴσην ἔχον ΑΒΓ accedilısı eşit olanangle ΑΒΓ τὴν ὑπὸ ΑΒΓ γωνίαν ΑΓΒ accedilısınato angle ΑΓΒ τῇ ὑπὸ ΑΓΒ γωνίᾳ

I say that λέγω ὅτι İddia ediyorum kialso side ΑΒ to side ΑΓ καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ΑΒ kenarı da ΑΓ kenarınais equal ἐστιν ἴση eşittir

For if unequal is ΑΒ to ΑΓ Εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ Ccediluumlnkuuml eğer ΑΒ eşit değil ise ΑΓ ke-one of them is greater ἡ ἑτέρα αὐτῶν μείζων ἐστίν narınaSuppose ΑΒ be greater ἔστω μείζων ἡ ΑΒ biri daha buumlyuumlktuumlrand there has been taken away καὶ ἀφῃρήσθω ΑΒ daha buumlyuumlk olan olsun

from the greater ΑΒ ἀπὸ τῆς μείζονος τῆς ΑΒ ve diyelim daha kuumlccediluumlk olan ΑΓ ke-to the less ΑΓ τῇ ἐλάττονι τῇ ΑΓ narına eşit olan ΔΒan equal ΔΒ ἴση ἡ ΔΒ daha buumlyuumlk olan ΑΒ kenarından ke-and there has been joined ΔΓ καὶ ἐπεζεύχθω ἡ ΔΓ silmiş olsun

ve ΔΓ birleştirilmiş olsun

Since then ΔΒ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ O zaman ΔΒ eşittir ΑΓ kenarınaand ΒΓ is common κοινὴ δὲ ἡ ΒΓ ve ΒΓ ortaktırso the two ΔΒ and ΒΓ δύο δὴ αἱ ΔΒ ΒΓ boumlylece ΔΒ ΒΓ ikilisi eşittirler ΑΓto the two ΑΓ and ΒΓ δύο ταῖς ΑΓ ΓΒ ΒΓ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΒΓ accedilısı eşittir ΑΓΒ accedilısınaand angle ΔΒΓ καὶ γωνία ἡ ὑπὸ ΔΒΓ dolayısıyla ΔΓ tabanı eşittir ΑΒ ta-to angle ΑΓΒ γωνίᾳ τῇ ὑπὸ ΑΓΒ banınais equal ἐστιν ἴση ve ΔΒΓ uumlccedilgeni eşit olacak ΑΓΒ uumlccedilge-therefore the base ΔΓ to the base ΑΒ βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ nineis equal ἴση ἐστίν daha kuumlccediluumlk daha buumlyuumlğeand triangle ΔΒΓ to triangle ΑΓΒ καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ki bu saccedilmadırwill be equal ἴσον ἔσται dolayısıyla ΑΒ değildir eşit değil ΑΓthe less to the greater τὸ ἔλασσον τῷ μείζονι kenarınawhich is absurd ὅπερ ἄτοπον dolayısıyla eşittirtherefore ΑΒ is not unequal to ΑΓ οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓtherefore it is equal ἴση ἄρα

If therefore in a triangle ᾿Εὰν τριγώνου Dolayısıyla eğer bir uumlccedilgenin birbirinetwo angles be equal to one another αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν eşit iki accedilısı varsaalso the sides that subtend the equal καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι eşit accedilıların goumlrduumlğuuml kenarlar eşittir

angles πλευραὶ mdashgoumlsterilmesi gereken tam buyduwill be equal to one another ἴσαι ἀλλήλαις ἔσονταιmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β Γ

Δ

On the same straight ᾿Επὶ τῆς αὐτῆς εὐθείας Aynı doğru uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι eşit iki başka doğru[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilmeyecekwill not be constructed οὐ συσταθήσονται bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις

For if possible Εἰ γὰρ δυνατόν Ccediluumlnkuuml eğer muumlmkuumlnseon the same straight ΑΒ ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ aynı ΑΒ doğrusunda

Literally lsquoanother and another pointrsquo more clearly in Englishlsquoto different pointsrsquo

In English as apparently in Greek parts can mean lsquoregionrsquomdashinthis case more precisely lsquosidersquo

According to Fowler ([ as p ] and [ as p ]) lsquoAs

is never to be regarded as a prepositionrsquo This is unfortunate sinceit means that the two constructions lsquoEqual to X rsquo and lsquoSame as X rsquoare not grammatically parallel (We have lsquoequal to himrsquo but lsquosameas hersquo) The constructions are parallel in Greek ἴσος + dative andαὐτός + dative

CHAPTER ELEMENTS

to two given straights ΑΓ ΓΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ ΓΒ verilmiş iki ΑΓ ΓΒ doğrusunatwo other straights ΑΔ ΔΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ ΔΒ eşit başka iki ΑΔ ΔΒ doğrusuequal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ mdashdiyelim inşa edilmiş olsunlarsuppose have been constructed συνεστάτωσαν bir ve başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ Γ ve ΔΓ and Δ τῷ τε Γ καὶ Δ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι şoumlyle ki ΓΑ eşit olmalı ΔΑ doğrusunaso that ΓΑ is equal to ΔΑ ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ aynı Α ucuna sahip olanhaving the same extremity as it Α τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α ve ΓΒ ΔΒ doğrusunaand ΓΒ to ΔΒ τὴν δὲ ΓΒ τῇ ΔΒ aynı Β ucuna sahip olanhaving the same extremity as it Β τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β ve ΓΔ birleştirilmiş olsunand suppose there has been joined καὶ ἐπεζεύχθωΓΔ ἡ ΓΔ

Because equal is ΑΓ to ΑΔ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ Ccediluumlnkuuml ΑΓ eşittir ΑΔ doğrusunaequal is ἴση ἐστὶ yine eşittiralso angle ΑΓΔ to ΑΔΓ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ ΑΓΔ ΑΔΓ accedilısınaGreater therefore [is] μείζων ἄρα dolayısıyla ΑΔΓ buumlyuumlktuumlr ΔΓΒΑΔΓ than ΔΓΒ ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ accedilısındanby much therefore [is] πολλῷ ἄρα dolayısıyla ΓΔΒ ccedilok daha buumlyuumlktuumlrΓΔΒ greater than ΔΓΒ ἡ ὑπὸ ΓΔΒ μείζων ἐστί τῆς ὑπὸ ΔΓΒ ΔΓΒ accedilısındanMoreover since equal is ΓΒ to ΔΒ πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ Uumlstelik ΓΒ eşit olduğu iccedilin ΔΒequal is also ἴση ἐστὶ καὶ doğrusunaangle ΓΔΒ to angle ΔΓΒ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ ΓΔΒ accedilısı eşittir ΔΓΒ accedilısınaBut it was also shown than it ἐδείχθη δὲ αὐτῆς καὶ Ama ondan ccedilok daha buumlyuumlk olduğumuch greater πολλῷ μείζων goumlsterilmiştiwhich is absurd ὅπερ ἐστὶν ἀδύνατον ki bu saccedilmadır

Not therefore Οὐκ ἄρα Şoumlyle olmaz dolayısıyla aynı doğruon the same straight ἐπὶ τῆς αὐτῆς εὐθείας uumlzerindeto the same two straights δύο ταῖς αὐταῖς εὐθείαις verilmiş iki doğruyatwo other straights ἄλλαι δύο εὐθεῖαι iki başka doğru eşit[which are] equal ἴσαι her biri birineeither to either ἑκατέρα ἑκατέρᾳ inşa edilecekwill be constructed συσταθήσονται başka bir noktayato one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ aynı taraftato the same parts ἐπὶ τὰ αὐτὰ μέρη aynı uccedilları olanhaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι başlangıccediltaki doğrularlaas the original lines ταῖς ἐξ ἀρχῆς εὐθείαις mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

ΔΓ

The Perseus Project Word Study Tool does not recognize συ-νεστάτωσαν here but it should be just the plural form of συνεστάτωwhich is used for example in Proposition I and which Perseus de-clares to be a passive perfect imperative The active third-personimperative ending -τωσαν (instead of the older -ντων) is said bySmyth [ ] to appear in prose after Thucydides This describesEuclid However I cannot explain from Smyth the use of an activeperfect (as opposed to aorist) form with passive meaning Presum-ably the verb is used lsquoimpersonallyrsquo The LSJ lexicon [] cites the

present proposition under συνίστημι See also the note at IThe Greek verb is an infinitive An infinitive clause may follow

ὥστε [ para p ] Compare the enunciation of Proposition Fowler ([ than p ] and [ than p ]) does grant

the possibility of construing lsquothanrsquo as a preposition though he dis-approves Then English cannot exactly mirror the Greek μείζων +genitive Turkish does mirror it with -den buumlyuumlk See note above

Here one must refer to the diagram

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenin varsa iki kenarı eşittwo sides τὰς δύο πλευρὰς olan iki kenarato two sides [ταῖς] δύο πλευραῖς her bir (kenar) birinehave equal ἴσας ἔχῃ ve varsa tabana eşit tabanıeither to either ἑκατέραν ἑκατέρᾳ ayrıca olacak accedilıya eşit accedilılarıand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην (yani) eşit kenarları goumlrenleralso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳthey will have equal ἴσην ἕξει[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶνsubtended περιεχομένην

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki uumlccedilgen ΑΒΓ ve ΔΕΖthe two sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki kenarı ΑΒ ΑΓ eşit olan ΔΕ ΔΖto the two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki kenarınınhaving equal ἴσας ἔχοντα her biri birineeither to either ἑκατέραν ἑκατέρᾳ ΑΒ ΔΕ kenarınaΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ve ΑΓ ΔΖ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve onlarınand let them have ἐχέτω δὲ ΒΓ tabanı eşit olsun ΕΖ tabanınabase ΒΓ equal to base ΕΖ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην

I say that λέγω ὅτι İddia ediyorum kialso angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dato angle ΕΔΖ γωνίᾳ τῇ ὑπὸ ΕΔΖ eşittir ΕΔΖ accedilısınais equal ἐστιν ἴση

For there being applied ᾿Εφαρμοζομένου γὰρ Ccediluumlnkuuml uumlstuumlne koyulursatriangle ΑΒΓ τοῦ ΑΒΓ τριγώνου ΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenininto triangle ΔΕΖ ἐπὶ τὸ ΔΕΖ τρίγωνον ve yerleştirilirseand there being placed καὶ τιθεμένου Β noktası Ε noktasınathe point Β on the point Ε τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον ve ΒΓ ΕΖ doğrusunaand the straight ΒΓ on ΕΖ τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ Γ noktası da yerleşecek Ζ noktasınaalso the point Γ will apply to Ζ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ sayesinde eşitliğinin ΒΓ doğrusununby the equality of ΒΓ to ΕΖ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ ΕΖ doğrusunaThen ΒΓ applying to ΕΖ ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ O zaman ΒΓ yerleştirilince ΕΖalso will apply ἐφαρμόσουσι καὶ doğrusunaΒΑ and ΓΑ to ΕΔ and ΔΖ αἱ ΒΑ ΓΑ ἐπὶ τὰς ΕΔ ΔΖ ΒΑ ve ΓΑ doğruları da yerleşeceklerFor if base ΒΓ to the base ΕΖ εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ΕΔ ve ΔΖ doğrularınaapply ἐφαρμόσει Ccediluumlnkuuml eğer ΒΓ yerleşirse ΕΖ ta-and sides ΒΑ ΑΓ to ΕΔ ΔΖ αἱ δὲ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ banınado not apply οὐκ ἐφαρμόσουσιν ve ΒΑ ΑΓ kenarları yerleşmezse ΕΔbut deviate ἀλλὰ παραλλάξουσιν ΔΖ kenarlarınaas ΕΗ ΗΖ ὡς αἱ ΕΗ ΗΖ ama saparsathere will be constructed συσταθήσονται ΕΗ ve ΗΖ olarakon the same straight ἐπὶ τῆς αὐτῆς εὐθείας inşa edilmiş olacakto two given straights δύο ταῖς αὐταῖς εὐθείαις aynı doğru uumlzerindetwo other straights equal ἄλλαι δύο εὐθεῖαι ἴσαι verilmiş iki doğruyaeither to either ἑκατέρα ἑκατέρᾳ iki başka doğru eşitto one and another point πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ her biri birineto the same parts ἐπὶ τὰ αὐτὰ μέρη başka bir noktayahaving the same extremities τὰ αὐτὰ πέρατα ἔχουσαι aynı taraftaBut they are not constructed οὐ συνίστανται δέ aynı uccedilları olantherefore it is not [the case] that οὐκ ἄρα Ama inşa edilmedilerthere being applied ἐφαρμοζομένης dolayısıyla (durum) şoumlyle değilthe base ΒΓ to the base ΕΖ τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν ΒΓ tabanı yerleştirilince ΕΖ tabanınathere do not apply οὐκ ἐφαρμόσουσι ΒΑ ΑΓ kenarları yerleşmez ΕΔ ΔΖsides ΒΑ ΑΓ to ΕΔ ΔΖ καὶ αἱ ΒΑ ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ ΔΖ kenarlarınaTherefore they apply ἐφαρμόσουσιν ἄρα Dolayısıyla yerleşirlerSo angle ΒΑΓ ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ Boumlylece ΒΑΓ accedilısı yerleşecek ΕΔΖ

CHAPTER ELEMENTS

to angle ΕΔΖ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ accedilısınawill apply ἐφαρμόσει ve ona eşit olacakand will be equal to it καὶ ἴση αὐτῇ ἔσται

If therefore two triangles ᾿Εὰν δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς varsa iki kenarıto two sides [ταῖς] δύο πλευραῖς eşit olanhave equal ἴσας ἔχῃ iki kenaraeither to either ἑκατέραν ἑκατέρᾳ her bir (kenar) birineand have also base equal to base ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην ve varsa tabana eşit tabanıalso angle to angle καὶ τὴν γωνίαν τῇ γωνίᾳ ayrıca olacak accedilıya eşit accedilılarıthey will have equal ἴσην ἕξει (yani) eşit kenarları goumlrenler[namely] that by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdashgoumlsterilmesi gereken tam buydusubtended περιεχομένηνmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α

Β

Γ

Δ

Ε

Η

Ζ

The given rectilineal angle Τὴν δοθεῖσαν γωνίαν εὐθύγραμμον Verilen duumlzkenar accedilıyıto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given rectilineal angle ἡ δοθεῖσα γωνία εὐθύγραμμος duumlzkenar bir accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ

Then it is necessary δεῖ δὴ Şimdi gereklidirto cut it in two αὐτὴν δίχα τεμεῖν onun ikiye kesilmesi

Suppose there has been chosen Εἰλήφθω Diyelim seccedililmiş olsunon ΑΒ at random a point Δ ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ ΑΒ uumlzerinde rastgele bir nokta Δand there has been taken from ΑΓ καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ ve kesilmiş olsun ΑΓ doğrusundanΑΕ equal to ΑΔ τῇ ΑΔ ἴση ἡ ΑΕ ΑΕ eşit olan ΑΔ doğrusunaand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand there has been constructed on ΔΕ καὶ συνεστάτω ἐπὶ τῆς ΔΕ ve inşa edilmiş olsun ΔΕ uumlzerindean equilateral triangle ΔΕΖ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ bir eşkenar uumlccedilgen ΔΕΖand ΑΖ has been joined καὶ ἐπεζεύχθω ἡ ΑΖ ve ΑΖ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kiangle ΒΑΓ has been cut in two ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ΒΑΓ accedilısı ikiye kesilmiş olduby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας ΑΖ doğrusu tarafındanFor because ΑΔ is equal to ΑΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ Ccediluumlnkuuml olduğundan ΑΔ eşit ΑΕ ke-and ΑΖ is common κοινὴ δὲ ἡ ΑΖ narınathen the two ΔΑ and ΑΖ δύο δὴ αἱ ΔΑ ΑΖ ve ΑΖ ortakto the two ΕΑ and ΑΖ δυσὶ ταῖς ΕΑ ΑΖ ΔΑ ΑΖ ikilisi eşittirler ΕΑ ΑΖ ikil-are equal ἴσαι εἰσὶν isinin

Here the generic article (see note to Proposition above) isparticularly appropriate Suppose we take a straight line with apoint A on it and draw a circle with center A cutting the line atB and C Then the straight line BC has been bisected at A Inparticular a line has been bisected But this does not mean we havesolved the problem of the present proposition In modern mathemat-

ical English the proposition could indeed be lsquoTo bisect a rectilinealanglersquo but then lsquoarsquo must be understood as lsquoan arbitraryrsquo or lsquoa givenrsquoOf course Euclid does supply this qualification in any case

For lsquocut in tworsquo we could say lsquobisectrsquo but in at least one placein Proposition δίχα τεμεῖν will be separated

either to either ἑκατέρα ἑκατέρᾳ her biri birine and the base ΔΖ to the base ΕΖ καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ve ΔΖ tabanı ΕΖ tabanına eşittiris equal ἴση ἐστίν dolayısıyla ΔΑΖ accedilısı ΕΑΖ accedilısınatherefore angle ΔΑΖ γωνία ἄρα ἡ ὑπὸ ΔΑΖ eşittirto angle ΕΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖis equal ἴση ἐστίν

Therefore the given rectilineal angle ῾Η ἄρα δοθεῖσα γωνία εὐθύγραμμος Dolayısıyla verilen duumlzkenar accedilı ΒΑΓΒΑΓ ἡ ὑπὸ ΒΑΓ kesilmiş oldu ikiyehas been cut in two δίχα τέτμηται ΑΖ doğrusuncaby the straight ΑΖ ὑπὸ τῆς ΑΖ εὐθείας mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Β

Α

Γ

Δ Ε

Ζ

The given bounded straight Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην Verilen sınırlı doğruyuto cut in two δίχα τεμεῖν ikiye kesmek

Let be ῎Εστω Verilmiş olsunthe given bounded straight line ΑΒ ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ bir sınırlı doğru ΑΒ

It is necessary then δεῖ δὴ Gereklidirthe bounded straight line ΑΒ to cut τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα verilmiş ΑΒ sınırlı doğrusunu kesmek

in two τεμεῖν ikiye

Suppose there has been constructed Συνεστάτω ἐπ᾿ αὐτῆς Kabul edelim ki uumlzerinde inşa edilmişon it τρίγωνον ἰσόπλευρον τὸ ΑΒΓ olsunan equilateral triangle ΑΒΓ καὶ τετμήσθω bir eşkenar uumlccedilgen ΑΒΓand suppose has been cut in two ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ ve ΑΓΒ accedilısı kesilmiş olsun ikiyethe angle ΑΓΒ by the straight ΓΔ ΓΔ doğrusunca

I say that λέγω ὅτι İddia ediyorum kithe straight ΑΒ has been cut in two ἡ ΑΒ εὐθεῖα δίχα τέτμηται ΑΒ doğrusu ikiye kesilmiş olduat the point Δ κατὰ τὸ Δ σημεῖον Δ noktasında Ccediluumlnkuuml ΑΓ eşitFor because ΑΓ is equal to ΑΒ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ olduğundan ΑΒ kenarınaand ΓΔ is common κοινὴ δὲ ἡ ΓΔ ve ΓΔ ortakthe two ΑΓ and ΓΔ δύο δὴ αἱ ΑΓ ΓΔ ΑΓ ve ΓΔ ikilisi eşittirler ΒΓ ΓΔ ik-to the two ΒΓ ΓΔ δύο ταῖς ΒΓ ΓΔ ilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΓΔ accedilısı eşittir ΒΓΔ accedilısınaand angle ΑΓΔ καὶ γωνία ἡ ὑπὸ ΑΓΔ dolayısıyla ΑΔ tabanı ΒΔ tabanınato angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ eşittiris equal ἴση ἐστίνtherefore the base ΑΔ to the base ΒΔ βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔis equal ἴση ἐστίν

Therefore the given bounded ῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένηstraight ἡ ΑΒ Dolayısıyla verilmiş sınırlı ΑΒ

ΑΒ δίχα τέτμηται κατὰ τὸ Δ doğrusuhas been cut in two at Δ ὅπερ ἔδει ποιῆσαι Δ noktasında ikiye kesilmiş oldumdashjust what it was necessary to do mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

Α Β

Γ

Δ

To the given straight Τῇ δοθείσῃ εὐθείᾳ Verilen bir doğruyafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilen bir noktadaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ bir doğru ΑΒand the given point on it Γ τὸ δὲ δοθὲν σημεῖον ἐπ᾿ αὐτῆς τὸ Γ ve uumlzerinde bir nokta Γ

It is necessary then δεῖ δὴ Gereklidirfrom the point Γ ἀπὸ τοῦ Γ σημείου Γ noktasındato the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusunaat right angles πρὸς ὀρθὰς γωνίας dik accedilılardato draw a straight line εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru

Suppose there has been chosen Εἰλήφθω Kabul edelim ki seccedililmiş olsunon ΑΓ at random a point Δ ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ ΑΓ doğrusunda rastgele bir nokta Δand there has been laid down καὶ κείσθω ve yerleştirilmiş olsuban equal to ΓΔ [namely] ΓΕ τῇ ΓΔ ἴση ἡ ΓΕ ΓΕ eşit olarak ΓΔ doğrusunaand there has been constructed καὶ συνεστάτω ve inşa edilmiş olsunon ΔΕ ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον ΔΕ uumlzerinde bir eşkenar uumlccedilgen ΖΔΕan equilateral triangle ΖΔΕ τὸ ΖΔΕ ve ΖΓ birleştirilmiş olsunand there has been joined ΖΓ καὶ ἐπεζεύχθω ἡ ΖΓ

I say that λέγω ὅτι İddia ediyorum kito the given straight line ΑΒ τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerindeki Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΖΓ doğrusu ccedilizilmişolduhas been drawn a straight line ΖΓ εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ Ccediluumlnkuuml ΔΓ eşit olduğundan ΓΕFor since ΔΓ is equal to ΓΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ doğrusunaand ΓΖ is common κοινὴ δὲ ἡ ΓΖ ve ΓΖ ortak olduğundanthe two ΔΓ and ΓΖ δύο δὴ αἱ ΔΓ ΓΖ ΔΓ ve ΓΖ ikilisito the two ΕΓ and ΓΖ δυσὶ ταῖς ΕΓ ΓΖ eşittirler ΕΓ ve ΓΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΔΖ tabanı eşittir ΖΕ tabanınaand the base ΔΖ to the base ΖΕ καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ dolayısıyla ΔΓΖ accedilısı eşittir ΕΓΖis equal ἴση ἐστίν accedilısınatherefore angle ΔΓΖ γωνία ἄρα ἡ ὑπὸ ΔΓΖ ve bitişiktirlerto angle ΕΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ Ne zaman bir doğruis equal ἴση ἐστίν bir doğru uumlzerinde dikilenand they are adjacent καί εἰσιν ἐφεξῆς bitişik accedilıları birbirine eşit yaparsaWhenever a straight ὅταν δὲ εὐθεῖα bu accedilıların her biri dik olurstanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα Dolayısıyla ΔΓΖ ΖΓΕ accedilılarının herthe adjacent angles τὰς ἐφεξῆς γωνίας ikisi de diktirequal to one another ἴσας ἀλλήλαιςmake ποιῇ

This is the first time among the propositions that Euclid writesout straight line (εὐθεῖα γραμμὴ) and not just straight (εὐθεῖα)

Literally lsquolead conductrsquo

either of the equal angles is right ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστινRight therefore is either of the angles ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶνΔΓΖ and ΖΓΕ ὑπὸ ΔΓΖ ΖΓΕ

Therefore to the given straight ΑΒ Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ Dolayısıyla verilen ΑΒ doğrusunafrom the given point on it ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου uumlzerinde verilmiş Γ noktasındaΓ τοῦ Γ dik accedilılardaat right angles πρὸς ὀρθὰς γωνίας bir ΓΖ doğrusu ccedilizilmiş olduhas been drawn the straight line ΓΖ εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ mdashyapılması gereken tam buydumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Ζ

Α ΒΔ Γ Ε

To the given unbounded straight ᾿Επὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον Verilen sınırlanmamış doğruyafrom the given point ἀπὸ τοῦ δοθέντος σημείου verilen bir noktadanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayanto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν bir dik doğru ccedilizmek

Let be ῎Εστω Verilmiş olsunthe given unbounded straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ bir sınırlanmamış doğru ΑΒand the given point τὸ δὲ δοθὲν σημεῖον ve bir noktawhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayan ΓΓ τὸ Γ

It is necessary then δεῖ δὴ Gereklidirto the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilmiş ΑΒ sınırlanmamış doğrusunaΑΒ τὴν ΑΒ verilmiş Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς bir dik doğru ccedilizmekto draw a perpendicular straight line κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν

For suppose there has been chosen Εἰλήφθω γὰρ Ccediluumlnkuuml kabul edelim ki seccedililmiş olsunon the other parts ἐπὶ τὰ ἕτερα μέρη ΑΒ doğrusunun diğer tarafındaof the straight ΑΒ τῆς ΑΒ εὐθείας rastgele bir Δ noktasıat random a point Δ τυχὸν σημεῖον τὸ Δ ve Γ merkezindeand to the center Γ καὶ κέντρῳ μὲν τῷ Γ ΓΔ uzaklığındaat the distance ΓΔ διαστήματι δὲ τῷ ΓΔ bir ccedilember ccedilizilmiş olsun ΕΖΗa circle has been drawn ΕΖΗ κύκλος γεγράφθω ὁ ΕΖΗ ve ΕΗ doğrusu Θ noktasında ikiye ke-and has been cut καὶ τετμήσθω silmiş olsunthe straight ΕΗ ἡ ΕΗ εὐθεῖα ve birleştirilmiş olsunin two at Θ δίχα κατὰ τὸ Θ ΓΗ ΓΘ ve ΓΕ doğrularıand there have been joined καὶ ἐπεζεύχθωσανthe straights ΓΗ ΓΘ and ΓΕ αἱ ΓΗ ΓΘ ΓΕ εὐθεῖαι

I say that λέγω ὅτι İddia ediyorum kito the given unbounded straight ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον verilen sınırlanmamış ΑΒ doğrusunaΑΒ τὴν ΑΒ verilen Γ noktasındanfrom the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ uumlzerinde olmayanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς ccedilizilmiş oldu dik ΓΘ doğrusuhas been drawn a perpendicular ΓΘ κάθετος ἦκται ἡ ΓΘ Ccediluumlnkuuml ΗΘ eşit olduğundan ΘΕFor because ΗΘ is equal to ΘΕ ᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ doğrusunaand ΘΓ is common κοινὴ δὲ ἡ ΘΓ ve ΘΓ ortakthe two ΗΘ and ΘΓ δύο δὴ αἱ ΗΘ ΘΓ ΗΘ ve ΘΓ ikilisi

CHAPTER ELEMENTS

to the two ΕΘ and ΘΓ are equal δύο ταῖς ΕΘ ΘΓ ἴσαι εἱσὶν eşittirler ΕΘ ve ΘΓ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΓΗ to the base ΓΕ καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ve ΓΗ tabanı eşittir ΓΕ tabanınais equal ἐστιν ἴση dolayısıyla ΓΘΗ accedilısı eşittir ΕΘΓtherefore angle ΓΘΗ γωνία ἄρα ἡ ὑπὸ ΓΘΗ accedilısınato angle ΕΘΓ γωνίᾳ τῇ ὑπὸ ΕΘΓ Ve onlar bitişiktirleris equal ἐστιν ἴση Ne zaman bir doğruand they are adjacent καί εἰσιν ἐφεξῆς bir doğru uumlzerinde dikildiğindeWhenever a straight ὅταν δὲ εὐθεῖα bitişik accedilıları birbirine eşit yaparsastanding on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα accedilıların her biri eşittirthe adjacent angles τὰς ἐφεξῆς γωνίας ve dikiltilen doğruequal to one another make ἴσας ἀλλήλαις ποιῇ uumlzerinderight ὀρθὴ dikildiği doğruya diktir denireither of the equal angles is ἑκατέρα τῶν ἴσων γωνιῶν ἐστινand καὶthe straight that has been stood ἡ ἐφεστηκυῖα εὐθεῖαis called perpendicular κάθετος καλεῖταιto that on which it has been stood ἐφ᾿ ἣν ἐφέστηκεν

Therefore to the given unbounded ᾿Επὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον Dolayısıyla verilen ΑΒ sınırlandırıl-straight ΑΒ τὴν ΑΒ mamış doğruya

from the given point Γ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ verilen Γ noktasındanwhich is not on it ὃ μή ἐστιν ἐπ᾿ αὐτῆς uumlzerinde olmayana perpendicular ΓΘ has been drawn κάθετος ἦκται ἡ ΓΘ bir dik ΓΘ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α Β

Γ

Δ

Ε

Ζ

Η Θ

If a straight ᾿Εὰν εὐθεῖα Eğer bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make [them] ποιήσει yapacak (onları)

For some straight ΑΒ Εὐθεῖα γάρ τις ἡ ΑΒ Ccediluumlnkuuml bir ΑΒ doğrusudastood on the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα dikiltilmiş ΓΔ doğrusumdashsuppose it makes angles γωνίας ποιείτω mdashkabul edelim ki ΓΒΑ ve ΑΒΔΓΒΑ and ΑΒΔ τὰς ὑπὸ ΓΒΑ ΑΒΔ accedilılarını oluşturmuş olsun

I say that λὲγω ὅτι İddia ediyorum kithe angles ΓΒΑ and ΑΒΔ αἱ ὑπὸ ΓΒΑ ΑΒΔ γωνίαι ΓΒΑ ve ΑΒΔ accedilılarıeither are two rights ἤτοι δύο ὀρθαί εἰσιν ya iki dik accedilıdıror [are] equal to two rights ἢ δυσὶν ὀρθαῖς ἴσαι ya da iki dik accedilıya eşittir(ler)

If equal is Εἰ μὲν οὖν ἴση ἐστὶν Eğer ΓΒΑ eşitse ΑΒΔ accedilısınaΓΒΑ to ΑΒΔ ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ iki dik accedilıdırlarthey are two rights δύο ὀρθαί εἰσιν

If not εἰ δὲ οὔ Eğer değilse

Euclid uses a present active imperative here

suppose there has been drawn ἤχθω kabul edelim ki ccedilizilmiş olsunfrom the point Β ἀπὸ τοῦ Β σημείου Β noktasındanto the [straight] ΓΔ τῇ ΓΔ [εὐθείᾳ] ΓΔ doğrusunaat right angles πρὸς ὀρθὰς dik accedilılardaΒΕ ἡ ΒΕ ΒΕ

Therefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔ iki diktirare two rights δύο ὀρθαί εἰσιν ve olduğundan ΓΒΕand since ΓΒΕ καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ eşit ΓΒΑ ve ΑΒΕ ikilisineto the two ΓΒΑ and ΑΒΕ is equal δυσὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ἴση ἐστίν ΕΒΔ her birine eklenmiş olsunlet there be added in common ΕΒΔ κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ Dolayısıyla ΓΒΕ ve ΕΒΔTherefore ΓΒΕ and ΕΒΔ αἱ ἄρα ὑπὸ ΓΒΕ ΕΒΔ eşittirlerto the three ΓΒΑ ΑΒΕ and ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ ΑΒΕ ΕΒΔ ΓΒΑ ΑΒΕ ve ΕΒΔ uumlccedilluumlsuumlneare equal ἴσαι εἰσίνMoreover πάλιν Dahasısince ΔΒΑ ἐπεὶ ἡ ὑπὸ ΔΒΑ olduğundan ΔΒΑto the two ΔΒΕ and ΕΒΑ is equal δυσὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ἴση ἐστίν eşit ΔΒΕ ve ΕΒΑ ikilisinelet there be added in common ΑΒΓ κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ ΑΒΓ her birine eklenmiş olsuntherefore ΔΒΑ and ΑΒΓ αἱ ἄρα ὑπὸ ΔΒΑ ΑΒΓ dolayısıyla ΔΒΑ ve ΑΒΓto the three ΔΒΕ ΕΒΑ and ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ ΕΒΑ ΑΒΓ eşittirlerare equal ἴσαι εἰσίν ΔΒΕ ΕΒΑ ve ΑΒΓ uumlccedilluumlsuumlneAnd ΓΒΕ and ΕΒΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔequal to the same three τρισὶ ταῖς αὐταῖς ἴσαι Ve ΓΒΕ ve ΕΒΔ accedilılarının goumlsteril-And equals to the same τὰ δὲ τῷ αὐτῷ ἴσα miştiare also equal to one another καὶ ἀλλήλοις ἐστὶν ἴσα eşitliği aynı uumlccedilluumlyealso therefore ΓΒΕ and ΕΒΔ καὶ αἱ ὑπὸ ΓΒΕ ΕΒΔ ἄρα Ve aynı şeye eşit olanlar birbirine eşit-to ΔΒΑ and ΑΒΓ are equal ταῖς ὑπὸ ΔΒΑ ΑΒΓ ἴσαι εἰσίν tirbut ΓΒΕ and ΕΒΔ ἀλλὰ αἱ ὑπὸ ΓΒΕ ΕΒΔ ve dolayısıyla ΓΒΕ ve ΕΒΔare two rights δύο ὀρθαί εἰσιν eşittirler ΔΒΑ ve ΑΒΓ accedilılarınaand therefore ΔΒΑ and ΑΒΓ καὶ αἱ ὑπὸ ΔΒΑ ΑΒΓ ἄρα ama ΓΒΕ veΕΒΔ iki diktirare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν ve dolayısıyla ΔΒΑ ve ΑΒΓ

iki dike eşittirler

If therefore a straight ᾿Εὰν ἄρα εὐθεῖα Eğer dolayısıyla bir doğrustood on a straight ἐπ᾿ εὐθεῖαν σταθεῖσα dikiltilmiş bir doğrunun uumlzerinemake angles γωνίας ποιῇ yaparsa accedilılareither two rights ἤτοι δύο ὀρθὰς ya iki dikor equal to two rights ἢ δυσὶν ὀρθαῖς ἴσας ya da iki dike eşitit will make ποιήσει [τηεμ] olacak (onları)mdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Ε

Δ

If to some straight ᾿Εὰν πρός τινι εὐθείᾳ Eğer bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıto two rights δυσὶν ὀρθαῖς ἴσας yaparsamake equal ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular

CHAPTER ELEMENTS

For to some straight ΑΒ Πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ Bir ΑΒ doğrusunaand at the same point Β καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β ve bir Β noktasındatwo straights ΒΓ and ΒΔ δύο εὐθεῖαι αἱ ΒΓ ΒΔ aynı tarafında kalmayannot lying to the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι iki ΒΓ ve ΒΔ doğrularınınthe adjacent angles τὰς ἐφεξῆς γωνίας ΑΒΓ ve ΑΒΔΑΒΓ and ΑΒΔ τὰς ὑπὸ ΑΒΓ ΑΒΔ bitişik accedilılarının iki dik accedilıequal to two rights δύο ὀρθαῖς ἴσας mdasholduğu kabul edilsinmdashsuppose they make ποιείτωσαν

I say that λέγω ὅτι İddia ediyorum kion a straight ἐπ᾿ εὐθείας ΒΔ ile ΓΒ bir doğrudadırwith ΓΒ is ΒΔ ἐστὶ τῇ ΓΒ ἡ ΒΔ

For if it is not Εἰ γὰρ μή ἐστι Ccediluumlnkuuml eğer değilsewith ΒΓ on a straight τῇ ΒΓ ἐπ᾿ εὐθείας bir doğruda ΒΓ ile[namely] ΒΔ ἡ ΒΔ ΒΔlet there be ἔστω olsunwith ΒΓ in a straight τῇ ΓΒ ἐπ᾿ εὐθείας bir doğruda ΒΓ ileΒΕ ἡ ΒΕ ΒΕ

For since the straight ΑΒ ᾿Επεὶ οὖν εὐθεῖα ἡ ΑΒ Ccediluumlnkuuml ΑΒ doğrusuhas stood to the straight ΓΒΕ ἐπ᾿ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν dikiltilmiş olur ΓΒΕ doğrusunatherefore angles ΑΒΓ and ΑΒΕ αἱ ἄρα ὑπὸ ΑΒΓ ΑΒΕ γωνίαι dolayısıyla ΑΒΓ ve ΑΒΕ accedilılarıare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAlso ΑΒΓ and ΑΒΔ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ ΑΒΔ Ayrıca ΑΒΓ ve ΑΒΔare equal to two rights δύο ὀρθαῖς ἴσαι eşittirler iki dik accedilıyaTherefore ΓΒΑ and ΑΒΕ αἱ ἄρα ὑπὸ ΓΒΑ ΑΒΕ Dolayısıyla ΓΒΑ ve ΑΒΕare equal to ΓΒΑ and ΑΒΔ ταῖς ὑπὸ ΓΒΑ ΑΒΔ ἴσαι εἰσίν eşittirler ΓΒΑ ve ΑΒΔ accedilılarınaIn common κοινὴ Ortak ΓΒΑ accedilısının ccedilıkartıldığı kabulsuppose there has been taken away ἀφῃρήσθω edilsinΓΒΑ ἡ ὑπὸ ΓΒΑ Dolayısıyla ΑΒΕ kalanıtherefore the remainder ΑΒΕ λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ eşittir ΑΒΔ kalanınato the remainder ΑΒΔ is equal λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση kuumlccediluumlk olan buumlyuumlğethe less to the greater ἡ ἐλάσσων τῇ μείζονι ki bu imkansızdırwhich is impossible ὅπερ ἐστὶν ἀδύνατον Dolayısıyla değildir [durum] şoumlyleTherefore it is not [the case that] οὐκ ἄρα ΒΕ bir doğrudadır ΓΒ doğrusuylaΒΕ is on a straight with ΓΒ ἐπ᾿ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ Benzer şekilde goumlstereceğiz kiSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι hiccedilbiri [oumlyledir] ΒΔ dışındano other [is so] except ΒΔ οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ Dolayısıyla ΓΒ bir doğrudadır ΒΔ ileTherefore on a straight ἐπ᾿ εὐθείας ἄραis ΓΒ with ΒΔ ἐστὶν ἡ ΓΒ τῇ ΒΔ

If therefore to some straight ᾿Εὰν ἄρα πρός τινι εὐθείᾳ Eğer dolayısıyla bir doğruyaand at the same point καὶ τῷ πρὸς αὐτῇ σημείῳ ve aynı noktasındatwo straights δύο εὐθεῖαι iki doğrunot lying in the same parts μὴ ἐπὶ αὐτὰ μέρη κείμεναι aynı tarafında kalmayanadjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarıtwo right angles δυσὶν ὀρθαῖς ἴσας yaparsamake ποιῶσιν iki dik accedilıya eşiton a straight ἐπ᾿ εὐθείας bir doğrudawill be with one another ἔσονται ἀλλήλαις olacaklar birbirleriylethe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α

ΒΓ Δ

Ε

The English perfect sounds strange here but the point may bethat the standing has already come to be and will continue

This seems to be the first use of the first person plural

If two straights cut one another ᾿Εὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας Eğer iki doğru keserse birbirinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit bir birine

For let the straights ΑΒ and ΓΔ Δύο γὰρ εὐθεῖαι αἱ ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğrularıcut one another τεμνέτωσαν ἀλλήλας kessinler birbirleriniat the point Ε κατὰ τὸ Ε σημεῖον Ε noktasında

I say that λέγω ὅτι İddia ediyorum kiequal are ἴση ἐστὶν eşittirlerangle ΑΕΓ to ΔΕΒ ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ ΑΕΓ accedilısı ΔΕΒ accedilısınaand ΓΕΒ to ΑΕΔ ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ ve ΓΕΒ accedilısı ΑΕΔ accedilısına

For since the straight ΑΕ ᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΕ Ccediluumlnkuuml ΑΕ doğrusuhas stood to the straight ΓΔ ἐπ᾿ εὐθεῖαν τὴν ΓΔ ἐφέστηκε yerleşmişti ΓΔ doğrusunamaking angles ΓΕΑ and ΑΕΔ γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ ΑΕΔ oluşturur ΓΕΑ ve ΑΕΔ accedilılarınıtherefore angles ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ γωνίαι dolayısıyla ΓΕΑ ve ΑΕΔ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaMoreover πάλιν Dahasısince the straight ΔΕ ἐπεὶ εὐθεῖα ἡ ΔΕ ΔΕ doğrusuhas stood to the straight ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΑΒ ἐφέστηκε dikiltilmişti ΑΒ doğrusunamaking angles ΑΕΔ and ΔΕΒ γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ ΔΕΒ oluşturarak ΑΕΔ ve ΔΕΒ accedilılarınıtherefore angles ΑΕΔ and ΔΕΒ αἱ ἄρα ὑπὸ ΑΕΔ ΔΕΒ γωνίαι dolayısıyla ΑΕΔ ve ΔΕΒ accedilılarıare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyaAnd ΓΕΑ and ΑΕΔ were shown ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ ΑΕΔ Ve ΓΕΑ ve ΑΕΔ accedilılarının goumlster-equal to two rights δυσὶν ὀρθαῖς ἴσαι ilmiştitherefore ΓΕΑ and ΑΕΔ αἱ ἄρα ὑπὸ ΓΕΑ ΑΕΔ eşitliği iki dik accedilıyaare equal to ΑΕΔ and ΔΕΒ ταῖς ὑπὸ ΑΕΔ ΔΕΒ ἴσαι εἰσίν dolayısıyla ΓΕΑ ve ΑΕΔIn common κοινὴ eşittirler ΑΕΔ ve ΔΕΒ accedilılarınasuppose there has been taken away ἀφῃρήσθω Ortak ΑΕΔ accedilısının ccedilıkartılmışΑΕΔ ἡ ὑπὸ ΑΕΔ olduğu kabul edilsintherefore the remainder ΓΕΑ λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ dolayısıyla ΓΕΑ kalanıis equal to the remainder ΒΕΔ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν eşittir ΒΕΔ kalanınasimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι benzer şekilde goumlsterilecek kialso ΓΕΒ and ΔΕΑ are equal καὶ αἱ ὑπὸ ΓΕΒ ΔΕΑ ἴσαι εἰσίν ΓΕΒ accedilısı da eşittir ΔΕΑ accedilısına

If therefore ᾿Εὰν ἄρα Eğer dolayısıylatwo straights cut one another δύο εὐθεῖαι τέμνωσιν ἀλλήλας iki doğru keserse bir birinithe vertical angles τὰς κατὰ κορυφὴν γωνίας ters accedilılarthey make equal to one another ἴσας ἀλλήλαις ποιοῦσιν oluşturlar eşit birbirinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

ΓΔ Ε

One of the sides of any triangle Παντὸς τριγώνου μιᾶς τῶν πλευρῶν Herhangi bir uumlccedilgenin kenarlarındanbeing extended προσεκβληθείσης birithe exterior angle ἡ ἐκτὸς γωνία uzatılıdığındathan either ἑκατέρας dış accedilı

The Greek is κατὰ κορυφὴν which might be translated as lsquoata headrsquo just as in the conclusion of I ΑΒ has been cut in twolsquoat Δrsquo κατὰ τὸ Δ But κορυφή and the Latin vertex can both meancrown of the head and in anatomical use the English vertical refers

to this crown Apollonius uses κορυφή for the vertex of a cone [pp ndash]

This is a rare moment when two things are said to be equalsimply and not equal to one another

CHAPTER ELEMENTS

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν her biris greater μείζων ἐστίν iccedil ve karşıt accedilıdan

buumlyuumlktuumlr

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeniand let there have been extended καὶ προσεκβεβλήσθω ve uzatılmış olsunits side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ onun ΒΓ kenarı Δ noktasına

I say that λὲγω ὅτι İddia ediyorum kithe exterior angle ΑΓΔ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ΑΓΔ dış accedilısıis greater μείζων ἐστὶν buumlyuumlktuumlrthan either ἑκατέρας her ikiof the two interior and opposite an- τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ ve ΒΑΓ iccedil ve karşıt accedilılarından

gles ΓΒΑ and ΒΑΓ ΓΒΑ ΒΑΓ γωνιῶν

Suppose ΑΓ has been cut in two at Ε Τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε ΑΓ kenarı E noktasından ikiye ke-and ΒΕ being joined καὶ ἐπιζευχθεῖσα ἡ ΒΕ silmiş olsunmdashsuppose it has been extended ἐκβεβλήσθω ve birleştirilen ΒΕon a straight to Ζ ἐπ᾿ εὐθείας ἐπὶ τὸ Ζ mdashuzatılmış olsunand there has been laid down καὶ κείσθω Ζ noktasına bir doğrudaequal to ΒΕ ΕΖ τῇ ΒΕ ἴση ἡ ΕΖ ve yerleştirilmiş olsunand there has been joined καὶ ἐπεζεύχθω ΒΕ doğrusuna eşit olan ΕΖΖΓ ἡ ΖΓ ve birleştirilmiş olsunand there has been drawn through καὶ διήχθω ΖΓΑΓ to Η ἡ ΑΓ ἐπὶ τὸ Η ve ccedilizilmiş olsun

ΑΓ doğrusu Η noktasına kadar

Since equal are ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΕ to ΕΓ ἡ μὲν ΑΕ τῇ ΕΓ ΑΕ ΕΓ doğrusunaand ΒΕ to ΕΖ ἡ δὲ ΒΕ τῇ ΕΖ ve ΒΕ ΕΖ doğrusunathe two ΑΕ and ΕΒ δύο δὴ αἱ ΑΕ ΕΒ ΑΕ ve ΕΒ ikilisito the two ΓΕ and ΕΖ δυσὶ ταῖς ΓΕ ΕΖ eşittirler ΓΕ ve ΕΖ ikilisininare equal ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ ve ΑΕΒ accedilısıand angle ΑΕΒ καὶ γωνία ἡ ὑπὸ ΑΕΒ eşittir ΖΕΓ accedilısınais equal to angle ΖΕΓ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν dikey olduklarındanfor they are vertical κατὰ κορυφὴν γάρ dolayısıyla ΑΒ tabanıtherefore the base ΑΒ βάσις ἄρα ἡ ΑΒ eşittir ΖΓ tabanınais equal to the base ΖΓ βάσει τῇ ΖΓ ἴση ἐστίν ve ΑΒΕ uumlccedilgeniand triangle ΑΒΕ καὶ τὸ ΑΒΕ τρίγωνον eşittir ΖΕΓ uumlccedilgenineis equal to triangle ΖΕΓ τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον ve kalan accedilılarand the remaining angles καὶ αἱ λοιπαὶ γωνίαι eşittirler kalan accedilılarınare equal to the remaining angles ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν her biri birineeither to either ἑκατέρα ἑκατέρᾳ (yani) eşit kenarları goumlrenlerwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν Dolayısıyla eşittirlerTherefore equal are ἴση ἄρα ἐστὶν ΕΓΔ ve ΕΓΖΒΑΕ and ΕΓΖ ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ Ama buumlyuumlktuumlrbut greater is μείζων δέ ἐστιν ΒΑΕ ΕΓΖ accedilısındanΕΓΔ than ΕΓΖ ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ dolayısıyla buumlyuumlktuumlrtherefore greater μείζων ἄρα ΑΓΔ ΒΑΕ accedilısından[is] ΑΓΔ than ΒΑΕ ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ Benzer şekildeSimilarly ῾Ομοίως δὴ ikiye kesilmiş olduğundan ΒΓ ΒΓ having been cut in two τῆς ΒΓ τετμημένης δίχα goumlsterilecek ki ΒΓΗit will be shown that ΒΓΗ δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ ΑΓΔ accedilısına eşit olanwhich is ΑΓΔ τουτέστιν ἡ ὑπὸ ΑΓΔ buumlyuumlktuumlr ΑΒΓ accedilısından[is] greater than ΑΒΓ μείζων καὶ τῆς ὑπὸ ΑΒΓ

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides μιᾶς τῶν πλευρῶν kenarlarından biribeing extended προσεκβληθείσης uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıthan either εκατέρας her bir

of the interior and opposite angles τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν iccedil ve karşıt accedilıdanis greater μείζων ἐστίν buumlyuumlktuumlrmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Ζ

Η

Two angles of any triangle Παντὸvς τριγώνου αἱ δύο γωνίαι Herhangi bir uumlccedilgenin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῇ μεταλαμβανόμεναι mdashnasıl alınırsa alınan

Let there be ῎Εστω Olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that ᾿λέγω ὅτι İddia ediyorum kitwo angles of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι ΑΒΓ uumlccedilgeninin iki accedilısıare less than two rights δύο ὀρθῶν ἐλάττονές εἰσι kuumlccediluumlktuumlr iki dik accedilıdanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl alınırsa alınsın

For suppose there has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml uzatılmış olsunΒΓ to Δ ἡ ΒΓ ἐπὶ τὸ Δ ΒΓ Δ noktasına

And since of triangle ΑΒΓ Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ Ve ΑΒΓ uumlccedilgenininΑΓΔ is an exterior angle ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ bir dış accedilısı olduğundan ΑΓΔit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΑΒΓ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ iccedil ve karşıt ΑΒΓ accedilısındanLet ΑΓΒ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ ΑΓΒ ortak accedilısı eklenmiş olsuntherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ dolayısıyla ΑΓΔ ve ΑΓΒare greater than ΑΒΓ and ΒΓΑ τῶν ὑπὸ ΑΒΓ ΒΓΑ μείζονές εἰσιν buumlyuumlktuumlrler ΑΒΓ ve ΒΓΑ accedilılarındanBut ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Ama ΑΓΔ ve ΑΓΒare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore ΑΒΓ and ΒΓΑ αἱ ἄρα ὑπὸ ΑΒΓ ΒΓΑ dolayısıyla ΑΒΓ ve ΒΓΑare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlrler iki dik accedilıdanSimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι Benzer şekilde goumlstereceğiz kialso ΒΑΓ and ΑΓΒ καὶ αἱ ὑπὸ ΒΑΓ ΑΓΒ ΒΑΓ ve ΑΓΒ deare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσι kuumlccediluumlktuumlrler iki dik accedilıdanand yet [so are] ΓΑΒ and ΑΒΓ καὶ ἔτι αἱ ὑπὸ ΓΑΒ ΑΒΓ ve sonra [oumlyledirler] ΓΑΒ ve ΑΒΓ

Therefore two angles of any triangle Παντὸvς ἄρα τριγώνου αἱ δύο γωνίαι Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than two rights δύο ὀρθῶν ἐλάσςονές εἰσι accedilısımdashtaken anyhow πάντῇ μεταλαμβανόμεναι kuumlccediluumlktuumlr iki dik accedilıdanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl alınırsa alınsın

mdashgoumlsterilmesi gereken tam buydu

Α

Β ΓΔ

CHAPTER ELEMENTS

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılar

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving side ΑΓ greater than ΑΒ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ ΑΓ kenarı daha buumlyuumlk olan ΑΒ ke-

narından

I say that λέγω ὅτι İddia ediyorum kialso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dais greater than ΒΓΑ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ daha buumlyuumlktuumlr ΒΓΑ accedilısından

For since ΑΓ is greater than ΑΒ ᾿Επεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ Ccediluumlnkuuml ΑΓ ΑΒ kenarından dahasuppose there has been laid down κείσθω buumlyuumlk olduğundanequal to ΑΒ τῇ ΑΒ ἴση yerleştirilmiş olsunΑΔ ἡ ΑΔ eşit olan ΑΒ kenarınaand let ΒΔ be joined καὶ ἐπεζεύχθω ἡ ΒΔ ΑΔ

ve ΒΔ birleştirilmiş olsun

Since also of triangle ΒΓΔ Καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ Ayrıca ΒΓΔ uumlccedilgenininangle ΑΔΒ is exterior ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ ΑΔΒ accedilısı dış accedilı olduğundanit is greater μείζων ἐστὶ buumlyuumlktuumlrthan the interior and opposite ΔΓΒ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ iccedil ve karşıt ΔΓΒ accedilısındanand ΑΔΒ is equal to ΑΒΔ ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ve ΑΔΒ eşittir ΑΒΔ accedilısınasince side ΑΒ is equal to ΑΔ ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση ΑΒ kenarı eşit olduğundan ΑΔ ke-greater therefore μείζων ἄρα narınais ΑΒΔ than ΑΓΒ καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr dolayısıylaby much therefore πολλῷ ἄρα ΑΒΔ ΑΓΒ accedilısındanΑΒΓ is greater ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ dolayısıyla ccedilok dahathan ΑΓΒ τῆς ὑπὸ ΑΓΒ buumlyuumlktuumlr ΑΒΓ

ΑΓΒ accedilısından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendethe greater side ἡ μείζων πλευρὰ daha buumlyuumlk bir kenarsubtends the greater angle τὴν μείζονα γωνίαν ὑποτείνει daha buumlyuumlk bir accedilıyı karşılarmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα daha buumlyuumlk bir accedilıthe greater side subtends γωνίαν ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanır

For let there be ῎Εστω Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgenihaving angle ΑΒΓ greater μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν ΑΒΓ accedilısı daha buumlyuumlk olanthan ΒΓΑ τῆς ὑπὸ ΒΓΑ ΒΓΑ accedilısından

This enunciation has almost the same words as that of the nextproposition The object of the verb ὑποτείνει is preceded by thepreposition ὑπό in the next enunciation and not here But the moreimportant difference would seem to be word order subject-object-

verb here and object-subject-verb in I This difference inorder ensures that I is the converse of I

Heath here uses the expedient of the passive lsquoThe greater angleis subtended by the greater sidersquo

I say that λέγω ὅτι İddia ediyorum kialso side ΑΓ καὶ πλευρὰ ἡ ΑΓ ΑΓ kenarı dais greater than side ΑΒ πλευρᾶς τῆς ΑΒ μείζων ἐστίν daha buumlyuumlktuumlr ΑΒ kenarından

For if not Εἰ γὰρ μή Ccediluumlnkuuml değil iseeither ΑΓ is equal to ΑΒ ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ya ΑΓ eşittir ΑΒ kenarınaor less ἢ ἐλάσσων ya da daha kuumlccediluumlktuumlr[but] ΑΓ is not equal to ΑΒ ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ (ama) ΑΓ eşit değildir ΑΒ kenarınafor [if it were] ἴση γὰρ ἂν ccediluumlnkuuml (eğer olsaydı)also ΑΒΓ would be equal to ΑΓΒ ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ ΑΒΓ da eşit olurdu ΑΓΒ accedilısınabut it is not οὐκ ἔστι δέ ama değildirtherefore ΑΓ is not equal to ΑΒ οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ dolayısıyla ΑΓ eşit değildir ΑΒ ke-Nor is ΑΓ less than ΑΒ οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ narınafor [if it were] ἐλάσσων γὰρ ΑΓ kuumlccediluumlk de değildir ΑΒ kenarındanalso angle ΑΒΓ would be [less] ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ ccediluumlnkuuml (eğer olsaydı)than ΑΓΒ τῆς ὑπὸ ΑΓΒ ΑΒΓ accedilısı da olurdu (kuumlccediluumlk)but it is not οὐκ ἔστι δέ ΑΓΒ accedilısındantherefore ΑΓ is not less than ΑΒ οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ ama değildirAnd it was shown that ἐδείχθη δέ ὅτι dolayısıyla ΑΓ kuumlccediluumlk değildir ΑΒ ke-it is not equal οὐδὲ ἴση ἐστίν narındanTherefore ΑΓ is greater than ΑΒ μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ Ve goumlsterilmişti ki

eşit değildirDolayısıyla ΑΓ daha buumlyuumlktuumlr ΑΒ ke-

narından

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgendeunder the greater angle ὑπὸ τὴν μείζονα γωνίαν daha buumlyuumlk bir accedilıthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk bir kenarca karşılanırmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β

Γ

Two sides of any triangle Παντὸς τριγώνου αἱ δύο πλευραὶ Herhangi bir uumlccedilgenin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsin

For let there be ῎Εστω γὰρ Ccediluumlnkuuml verilmiş olsuna triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ bir ΑΒΓ uumlccedilgeni

I say that λέγω ὅτι İddia ediyorum kitwo sides of triangle ΑΒΓ τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ ΑΒΓ uumlccedilgeninin iki kenarıare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι daha buumlyuumlktuumlr geriye kalandanmdashtaken anyhow πάντῃ μεταλαμβανόμεναι mdashnasıl seccedililirse seccedililsinΒΑ and ΑΓ than ΒΓ αἱ μὲν ΒΑ ΑΓ τῆς ΒΓ ΒΑ ve ΑΓ ΒΓ kenarındanΑΒ and ΒΓ than ΑΓ αἱ δὲ ΑΒ ΒΓ τῆς ΑΓ ΑΒ ve ΒΓ ΑΓ kenarındanΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΒΓ ve ΓΑ ΑΒ kenarından

For suppose has been drawn through Διήχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunΒΑ to a point Δ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον ΒΑ kenarı geccedilerek bir Δ noktasından

Literally lsquowasrsquo but this conditional use of was is archaic in En- glish

CHAPTER ELEMENTS

and there has been laid down καὶ κείσθω ve yerleştirilmiş olsunΑΔ equal to ΓΑ τῇ ΓΑ ἴση ἡ ΑΔ ΑΔ ΓΑ kenarına eşit olanand there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΔΓ ἡ ΔΓ ΔΓ

Since ΔΑ is equal to ΑΓ ᾿Επεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ ΔΑ eşit olduğundan ΑΓ kenarınaequal also is ἴση ἐστὶ καὶ eşittir ayrıcaangle ΑΔΓ to ΑΓΔ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ ΑΔΓ ΑΓΔ accedilısınaTherefore ΒΓΔ is greater than ΑΔΓ μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ Dolayısıyla ΒΓΔ buumlyuumlktuumlr ΑΔΓalso since there is a triangle ΔΓΒ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ accedilısındanhaving angle ΓΒΔ greater μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν yine ΔΓΒ bir uumlccedilgen olduğundanthan ΔΒΓ τῆς ὑπὸ ΒΔΓ ΒΓΔ daha buumlyuumlk olanand under the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ΒΔΓ accedilısındanthe greater side subtends ἡ μείζων πλευρὰ ὑποτείνει daha buumlyuumlk accedilıtherefore ΔΒ is greater than ΒΓ ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων daha buumlyuumlk kenarca karşılandışındanBut ΔΑ is equal to ΑΓ ἴση δὲ ἡ ΔΑ τῇ ΑΓ dolayısıyla ΔΒ buumlyuumlktuumlr ΒΓ kenarın-therefore ΒΑ and ΑΓ are greater μείζονες ἄρα αἱ ΒΑ ΑΓ danthan ΒΓ τῆς ΒΓ Ama ΔΑ eşittir ΑΓ kenarınasimilarly we shall show that ὁμοίως δὴ δείξομεν ὅτι dolayısıyla ΒΑ ve ΑΓ buumlyuumlktuumlrlerΑΒ and ΒΓ than ΓΑ καὶ αἱ μὲν ΑΒ ΒΓ τῆς ΓΑ ΒΓ kenarındanare greater μείζονές εἰσιν benzer şekilde goumlstereceğiz kiand ΒΓ and ΓΑ than ΑΒ αἱ δὲ ΒΓ ΓΑ τῆς ΑΒ ΑΒ ve ΒΓ ΓΑ kenarından

buumlyuumlktuumlrlerve ΒΓ ve ΓΑ ΑΒ kenarından

Therefore two sides of any triangle Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ Dolayısıyla herhangi bir uumlccedilgenin ikiare greater than the remaining one τῆς λοιπῆς μείζονές εἰσι kenarımdashtaken anyhow πάντῃ μεταλαμβανόμεναι daha buumlyuumlktuumlr geriye kalandanmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashnasıl seccedililirse seccedililsin

mdashgoumlsterilmesi gereken tam buydu

Ζ

Α

Β

Δ

Γ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendeon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

triangle ἐλάττονες μὲν ἔσονται daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέξουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle

For of a triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininon one of the sides ΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ bir ΒΓ kenarınınfrom its extremities Β and Γ ἀπὸ τῶν περάτων τῶν Β Γ Β ve Γ uccedillarındansuppose two straights have been δύο εὐθεῖαι ἐντὸς συνεστάτωσαν iccedileride iki doğru inşa edilmiş olsun

constructed within αἱ ΒΔ ΔΓ ΒΔ ve ΔΓΒΔ and ΔΓ

Heathrsquos version is lsquoSince DCB [ΔΓΒ] is a triangle rsquoHere the Greek verb συνίστημι is the same one used in I for

the contruction of a triangle on a given straight line Is it supposed

to be obvious to the reader even without a diagram that now thetwo constructed straight lines are supposed to meet at a point Seealso I and note

I say that λέγω ὅτι İddia ediyorum kiΒΔ and ΔΓ αἱ ΒΔ ΔΓ ΒΔ ve ΔΓthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki

triangle τῶν ΒΑ ΑΓ ΒΑ ve ΑΓ kenarındanΒΑ and ΑΓ ἐλάσσονες μέν εἰσιν daha kuumlccediluumltuumlrlerare less μείζονα δὲ γωνίαν περιέχουσι ama iccedilerirlerbut contain a greater angle τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ ΒΑΓ accedilısından daha buumlyuumlk ΒΔΓΒΔΓ than ΒΑΓ accedilısını

For let ΒΔ be drawn through to Ε Διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε Ccediluumlnkuuml ΒΔ ccedilizilmiş olsun Ε noktasınadoğru

And since of any triangle καὶ ἐπεὶ παντὸς τριγώνου Ve herhangi bir uumlccedilgenintwo sides than the remaining one αἱ δύο πλευραὶ τῆς λοιπῆς iki kenarı geriye kalandanare greater μείζονές εἰσιν buumlyuumlk olduğundanof the triangle ΑΒΕ τοῦ ΑΒΕ ἄρα τριγώνου ΑΒΕ uumlccedilgenininthe two sides ΑΒ and ΑΕ αἱ δύο πλευραὶ αἱ ΑΒ ΑΕ iki kenarı ΑΒ ve ΑΕare greater than ΒΕ τῆς ΒΕ μείζονές εἰσιν buumlyuumlktuumlr ΒΕ kenarındansuppose has been added in common κοινὴ προσκείσθω ortak olarak eklenmiş olsunΕΓ ἡ ΕΓ ΕΓtherefore ΒΑ and ΑΓ than ΒΕ and ΕΓ αἱ ἄρα ΒΑ ΑΓ τῶν ΒΕ ΕΓ dolayısıyla ΒΑ ve ΑΓ ΒΕ ve ΕΓ ke-are greater μείζονές εἰσιν narlarındanMoreover πάλιν buumlyuumlktuumlrlersince of the triangle ΓΕΔ ἐπεὶ τοῦ ΓΕΔ τριγώνου Dahasıthe two sides ΓΕ and ΕΔ αἱ δύο πλευραὶ αἱ ΓΕ ΕΔ ΓΕΔ uumlccedilgenininare greater than ΓΔ τῆς ΓΔ μείζονές εἰσιν iki kenarları ΓΕ ve ΕΔsuppose has been added in common κοινὴ προσκείσθω buumlyuumlktuumlr ΓΔ kenarındanΔΒ ἡ ΔΒ ortak olarak eklenmiş olsuntherefore ΓΕ and ΕΒ than ΓΔ andΔΒ αἱ ΓΕ ΕΒ ἄρα τῶν ΓΔ ΔΒ ΔΒare greater μείζονές εἰσιν dolayısıyla ΓΕ ve ΕΒ ΓΔ ve ΔΒ ke-But than ΒΕ and ΕΓ ἀλλὰ τῶν ΒΕ ΕΓ narlarındanΒΑ and ΑΓ were shown greater μείζονες ἐδείχθησαν αἱ ΒΑ ΑΓ buumlyuumlktuumlrlertherefore by much πολλῷ ἄρα Ama ΒΕ ve ΕΓ kenarlarındanΒΑ and ΑΓ than ΒΔ and ΔΓ αἱ ΒΑ ΑΓ τῶν ΒΔ ΔΓ ΒΑ ve ΑΓ kenarlarının goumlsterilmiştiare greater μείζονές εἰσιν buumlyuumlkluumlğuuml

dolayısıyla ccedilok daha buumlyuumlktuumlrΒΑ ve ΑΓ ΒΔ ve ΔΓ kenarlarından

Again Πάλιν Tekrarsince of any triangle ἐπεὶ παντὸς τριγώνου herhangi bir uumlccedilgeninthe external angle ἡ ἐκτὸς γωνία dış accedilısıthan the interior and opposite angle τῆς ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilısındanis greater μείζων ἐστίν daha buumlyuumlktuumlrtherefore of the triangle ΓΔΕ τοῦ ΓΔΕ ἄρα τριγώνου dolayısıyla ΓΔΕ uumlccedilgenininthe exterior angle ΒΔΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ dış accedilısı ΒΔΓis greater than ΓΕΔ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ buumlyuumlktuumlr ΓΕΔ accedilısındanFor the same [reason] again διὰ ταὐτὰ τοίνυν Aynı [sebepten] tekrarof the triangle ΑΒΕ καὶ τοῦ ΑΒΕ τριγώνου ΑΒΕ uumlccedilgenininthe exterior angle ΓΕΒ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ dış accedilısı ΓΕΒis greater than ΒΑΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΑΓ accedilısındanBut than ΓΕΒ ἀλλὰ τῆς ὑπὸ ΓΕΒ Ama ΓΕΒ accedilısındanΒΔΓ was shown greater μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ ΒΔΓ accedilısının buumlyuumlkluumlğuuml goumlsterilmiştitherefore by much πολλῷ ἄρα dolayısıyla ccedilok dahaΒΔΓ is greater than ΒΑΓ ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ buumlyuumlktuumlr ΒΔΓ ΒΑΓ accedilısından

If therefore of a triangle ᾿Εὰν ἄρα τριγώνου Eğer dolayısıyla bir uumlccedilgeninon one of the sides ἐπὶ μιᾶς τῶν πλευρῶν kenarlardan birininfrom its extremities ἀπὸ τῶν περάτων uccedillarındantwo straights δύο εὐθεῖαι iki doğrube constructed within ἐντὸς συσταθῶσιν iccedileride inşa edilirsethe constructed [straights] αἱ συσταθεῖσαι inşa edilen doğrularthan the remaining two sides of the τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν uumlccedilgenin geriye kalan iki kenarından

CHAPTER ELEMENTS

triangle ἐλάττονες μέν εἰσιν daha kuumlccediluumlk olacakwill be less μείζονα δὲ γωνίαν περιέχουσιν ama daha buumlyuumlk bir accedilıyı iccedilereceklerbut will contain the a greater angle ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show

Α

Β Γ

Δ

Ε

From three straights ᾿Εκ τριῶν εὐθειῶν Uumlccedil doğrudanwhich are equal αἵ εἰσιν ἴσαι eşit olanto three given [straights] τρισὶ ταῖς δοθείσαις [εὐθείαις] verilmiş uumlccedil doğruyaa triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgen oluşturulmasıand it is necessary δεῖ δὲ2 ve gereklidirfor two than the remaining one τὰς δύο τῆς λοιπῆς ikisinin kalandanto be greater μείζονας εἶναι daha buumlyuumlk olması[because of any triangle πάντῃ μεταλαμβανομένας (ccediluumlnkuuml herhangi bir uumlccedilgenintwo sides [διὰ τὸ καὶ παντὸς τριγώνου iki kenarıare greater than the remaining one τὰς δύο πλευρὰς buumlyuumlktuumlr geriye kalandantaken anyhow] τῆς λοιπῆς μείζονας εἶναι nasıl seccedililirse seccedililsin)

πάντῃ μεταλαμβανομένας]

Let be ῎Εστωσαν Verilmiş olsunthe given three straights αἱ δοθεῖσαι τρεῖς εὐθεῖαι uumlccedil doğruΑ Β and Γ αἱ Α Β Γ Α Β ve Γof which two than the remaining one ὧν αἱ δύο τῆς λοιπῆς ikisi kalandanare greater μείζονες ἔστωσαν buumlyuumlk olantaken anyhow πάντῃ μεταλαμβανόμεναι nasıl seccedililirse seccedililsinΑ and Β than Γ αἱ μὲν Α Β τῆς Γ Α ile Β Γ kenarındanΑ and Γ than Β αἱ δὲ Α Γ τῆς Β Α ile Γ Β kenarındanand Β and Γ than Α καὶ ἔτι αἱ Β Γ τῆς Α ve Β ile Γ Α kenarından

Is is necessary δεῖ δὴ Gereklidirfrom equals to Α Β and Γ ἐκ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olanlardanfor a triangle to be constructed τρίγωνον συστήσασθαι bir uumlccedilgenin inşa edilmesi

Suppose there is laid out ᾿Εκκείσθω Yerleştirilmiş olsunsome straight line ΔΕ τις εὐθεῖα ἡ ΔΕ bir ΔΕ doğrusubounded at Δ πεπερασμένη μὲν κατὰ τὸ Δ Δ noktasında sınırlanmışbut unbounded at Ε ἄπειρος δὲ κατὰ τὸ Ε ama Ε noktasında sınırlandırılmamışand there is laid down καὶ κείσθω yerleştirilmiş olsunΔΖ equal to Α τῇ μὲν Α ἴση ἡ ΔΖ Α doğrusuna eşit ΔΖΖΗ equal to Β τῇ δὲ Β ἴση ἡ ΖΗ Β doğrusuna eşit ΖΗand ΗΘ equal to Γ τῇ δὲ Γ ἴση ἡ ΗΘ ve Γ doğrusuna eşit ΗΘ and to center Ζ καὶ κέντρῳ μὲν τῷ Ζ ve Ζ merkezineat distance ΖΔ διαστήματι δὲ τῷ ΖΔ ΖΔ uzaklığındaa circle has been drawn ΔΚΛ κύκλος γεγράφθω ὁ ΔΚΛ bir ΔΚΛ ccedilemberi ccedilizilmiş olsunmoreover πάλιν dahasıto center Η κέντρῳ μὲν τῷ Η Η merkezineat distance ΗΘ διαστήματι δὲ τῷ ΗΘ ΗΘ uzaklığındacircle ΚΛΘ has been drawn κύκλος γεγράφθω ὁ ΚΛΘ ΚΛΘ ccedilemberi ccedilizilmiş olsun

In the Greek this is the infinitive εἶναι lsquoto bersquo as in the previousclause

According to Heiberg the manuscripts have δεῖ δή here as at

the beginnings of specifications (see sect) but Proclus and Eutociushave δεῖ δέ in their commentaries

and ΚΖ and ΚΗ have been joined καὶ ἐπεζεύχθωσαν αἱ ΚΖ ΚΗ ve ΚΖ ile ΚΗ birleştirilmiş olsun

I say that λέγω ὅτι İddia ediyorum kifrom three straights ἐκ τριῶν εὐθειῶν uumlccedil doğrudanequal to Α Β and Γ τῶν ἴσων ταῖς Α Β Γ Α Β ve Γ doğrularına eşit olana triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗ bir ΚΖΗ uumlccedilgeni inşa edilmiştir

For since the point Ζ ᾿Επεὶ γὰρ τὸ Ζ σημεῖον Ccediluumlnkuuml merkezi olduğundan Ζ noktasıis the center of circle ΔΚΛ κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου ΔΚΛ ccedilemberininΖΔ is equal to ΖΚ ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ ΖΔ eşittir ΖΚ doğrusunabut ΖΔ is equal to Α ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση ama ΖΔ eşittir Α doğrusunaAnd ΚΖ is therefore equal to Α καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση Ve ΚΖ dolayısıyla Α doğrusuna eşittirMoreover πάλιν Dahasısince the point Η ἐπεὶ τὸ Η σημεῖον merkezi olduğundan Η noktasıis the center of circle ΛΚΘ κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου ΛΚΘ ccedilemberininΗΘ is equal to ΗΚ ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ ΗΘ eşittir ΗΚ doğrusunabut ΗΘ is equal to Γ ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση ama ΗΘ eşittir Γ doğrusunaand ΚΗ is therefore equal to Γ καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση ve ΚΗ dolayısıyla Γ doğrusuna eşittirand ΖΗ is equal to Β ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση ve ΖΗ eşittir Β doğrusunatherefore the three straights αἱ τρεῖς ἄρα εὐθεῖαι dolayısıyla uumlccedil doğruΚΖ ΖΗ and ΗΚ αἱ ΚΖ ΖΗ ΗΚ ΚΖ ΖΗ ve ΗΚare equal to the three Α Β and Γ τρισὶ ταῖς Α Β Γ ἴσαι εἰσίν eşittirler Α Β ve Γ uumlccedilluumlsuumlne

Therefore from the three straights ᾿Εκ τριῶν ἄρα εὐθειῶνΚΖ ΖΗ and ΗΚ τῶν ΚΖ ΖΗ ΗΚwhich are equal αἵ εἰσιν ἴσαιto the three given straights τρισὶ ταῖς δοθείσαις εὐθείαιςΑ Β and Γ ταῖς Α Β Γa triangle has been constructed ΚΖΗ τρίγωνον συνέσταται τὸ ΚΖΗmdashjust what it was necessary to show ὅπερ ἔδει ποιῆσαι

Dolayısıyla uumlccedil doğrudanΚΖ ΖΗ ve ΗΚeşit olanverilmiş uumlccedil doğruyaΑ Β ve Γbir ΚΖΗ uumlccedilgeni inşa edilmiştirmdashgoumlsterilmesi gereken tam buydu

ΑΒΓ

Δ ΕΖ Η

Κ

Θ

Λ

At the given straight Πρὸς τῇ δοθείσῃ εὐθείᾳ Verilmiş bir doğrudaand at the given point on it καὶ τῷ πρὸς αὐτῇ σημείῳ ve uumlzerinde verilmiş noktadaequal to the given rectilineal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην verilmiş duumlzkenar accedilıya eşit olana rectilineal angle to be constructed γωνίαν εὐθύγραμμον συστήσασθαι bir duumlzkenar accedilı inşa edilmesi

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusuthe point on it Α τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α uumlzerindeki Α noktasıthe given rectilineal angle ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος verilmiş olsun duumlzkenar accedilıΔΓΕ ἡ ὑπὸ ΔΓΕ ΔΓΕ

It is necessary then δεῖ δὴ Gereklidir şimdi

CHAPTER ELEMENTS

on the given straight ΑΒ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ verilmiş ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ve uumlzerindeki Α noktasındato the given rectilileal angle τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilmiş duumlzkenarΔΓΕ τῇ ὑπὸ ΔΓΕ ΔΓΕ accedilısınaequal ἴσην eşitfor a rectilineal angle γωνίαν εὐθύγραμμον bir duumlzkenar accedilınınto be constructed συστήσασθαι inşa edilmesi

Suppose there have been chosen Εἰλήφθω Seccedililmiş olsunon either of ΓΔ and ΓΕ ἐφ᾿ ἑκατέρας τῶν ΓΔ ΓΕ ΓΔ ve ΓΕ doğrularının her birindenrandom points Δ and Ε τυχόντα σημεῖα τὰ Δ Ε rastgele Δ ve Ε noktalarıand ΔΕ has been joined καὶ ἐπεζεύχθω ἡ ΔΕ ve ΔΕ birleştirilmiş olsunand from three straights καὶ ἐκ τριῶν εὐθειῶν ve uumlccedil doğrudanwhich are equal to the three αἵ εἰσιν ἴσαι τρισὶ eşit olan verilmiş uumlccedilΓΔ ΔΕ and ΓΕ ταῖς ΓΔ ΔΕ ΓΕ ΓΔ ΔΕ ve ΓΕ doğrularınatriangle ΑΖΗ has been constructed τρίγωνον συνεστάτω τὸ ΑΖΗ bir ΑΖΗ uumlccedilgen inşa edilmiş olsunso that equal are ὥστε ἴσην εἶναι oumlyle ki eşit olsunΓΔ to ΑΖ τὴν μὲν ΓΔ τῇ ΑΖ ΓΔ ΑΖ doğrusunaΓΕ to ΑΗ τὴν δὲ ΓΕ τῇ ΑΗ ΓΕ ΑΗ doğrusuna ve ΔΕ ΖΗand ΔΕ to ΖΗ καὶ ἔτι τὴν ΔΕ τῇ ΖΗ doğrusuna

Since then the two ΔΓ and ΓΕ ᾿Επεὶ οὖν δύο αἱ ΔΓ ΓΕ O zaman ΔΓ ve ΓΕ ikilisiare equal to the two ΖΑ and ΑΗ δύο ταῖς ΖΑ ΑΗ ἴσαι εἰσὶν eşit olduğundan ΖΑ ve ΑΗ ikilisinineither to either ἑκατέρα ἑκατέρᾳ her biri birineand the base ΔΕ to the base ΖΗ καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ve ΔΕ tabanı ΖΗ tabanınais equal ἴση eşittherefore the angle ΔΓΕ γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ dolayısıyla ΔΓΕ accedilısıis equal to ΖΑΗ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση eşittir ΖΑΗ accedilısına

Therefore on the given straight Πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ DolayısıylaΑΒ τῇ ΑΒ ΑΒ doğrusundaand at the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ve uumlzerindeki Α noktasındaequal to the given rectilineal angle δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ verilen duumlzkenar ΔΓΕ accedilısına eşit

ΔΓΕ ΔΓΕ ἴση ΖΑΗ duumlzkenar accedilısı inşa edilmiştirthe rectilineal angle ΖΑΗ has been γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ mdashyapılması gereken tam buydu

constructed ΖΑΗmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι

Α Β

Γ

Δ

Ε

Ζ

Η

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides [ταῖς] δύο πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacak

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenimdashtwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ mdash iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınamdashand the angle at Α ἡ δὲ πρὸς τῷ Α γωνία mdashve Α noktasındaki accedilısıthan the angle at Δ τῆς πρὸς τῷ Δ γωνίας Δ doktasındakindenlet it be greater μείζων ἔστω buumlyuumlk olsun

I say that λέγω ὅτι İddia ediyorum kialso the base ΒΓ καὶ βάσις ἡ ΒΓ ΒΓ tabanı dathan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanis greater μείζων ἐστίν buumlyuumlktuumlr

For since [it] is greater ᾿Επεὶ γὰρ μείζων Ccediluumlnkuuml buumlyuumlk olduğundan[namely] angle ΒΑΓ ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısıthan angle ΕΔΖ τῆς ὑπὸ ΕΔΖ γωνίας ΕΔΖ accedilısındansuppose has been constructed συνεστάτω inşa edilmiş olsunon the straight ΔΕ πρὸς τῇ ΔΕ εὐθείᾳ ΔΕ doğrusundaand at the point Δ on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ ve uumlzerindeki Δ noktasındaequal to angle ΒΑΓ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ΒΑΓ accedilısına eşitΕΔΗ ἡ ὑπὸ ΕΔΗ ΕΔΗand suppose is laid down καὶ κείσθω ve yerleştirilmiş olsunto either of ΑΓ and ΔΖ equal ὁποτέρᾳ τῶν ΑΓ ΔΖ ἴση ΑΓ ve ΔΖ kenarlarının ikisine de eşitΔΗ ἡ ΔΗ ΔΗand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΕΗ and ΖΗ αἱ ΕΗ ΖΗ ΕΗ ve ΖΗ

Since [it] is equal ᾿Επεὶ οὖν ἴση ἐστὶν Eşit olduğundanΑΒ to ΔΕ ἡ μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΗ ἡ δὲ ΑΓ τῇ ΔΗ ve ΑΓ ΔΗ kenarınathe two ΒΑ and ΑΓ δύο δὴ αἱ ΒΑ ΑΓ ΒΑ ve ΑΓ ikilisito the two ΕΔ and ΔΗ δυσὶ ταῖς ΕΔ ΔΗ ΕΔ ve ΔΗ iklisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ve ΒΑΓ accedilısıto angle ΕΔΗ is equal γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση ΕΔΗ accedilısına eşittirtherefore the base ΒΓ βάσις ἄρα ἡ ΒΓ dolayısıyla ΒΓ tabanıto the base ΕΗ is equal βάσει τῇ ΕΗ ἐστιν ἴση ΕΗ tabanına eşittirMoreover πάλιν Dahasısince [it] is equal ἐπεὶ ἴση ἐστὶν eşit olduğundan[namely] ΔΖ to ΔΗ ἡ ΔΖ τῇ ΔΗ ΔΖ ΔΗ kenarına[it] too is equal ἴση ἐστὶ καὶ yine eşittir[namely] angle ΔΗΖ to ΔΖΗ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ ΔΗΖ accedilısı ΔΖΗ accedilısınatherefore [it] is greater μείζων ἄρα dolayısıyla buumlyuumlktuumlr[namely] ΔΖΗ than ΕΗΖ ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ ΔΖΗ ΕΗΖ accedilısındantherefore [it] is much greater πολλῷ ἄρα μείζων ἐστὶν dolayısıyla ccedilok daha buumlyuumlktuumlr[namely] ΕΖΗ than ΕΗΖ ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ ΕΖΗ ΕΗΖ accedilısındanAnd since there is a triangle ΕΖΗ καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ Ve ΕΖΗ bir uumlccedilgen olduğundanhaving greater μείζονα ἔχον buumlyuumlk olanangle ΕΖΗ than ΕΗΖ τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ ΕΖΗ accedilısı ΕΗΖ accedilısındanand the greater angle ὑπὸ δὲ τὴν μείζονα γωνίαν ve daha buumlyuumlk accedilımdashthe greater side subtends it ἡ μείζων πλευρὰ ὑποτείνει mdashdaha buumlyuumlk accedilı tarafından karşı-greater therefore also is μείζων ἄρα καὶ landığındanside ΕΗ than ΕΖ πλευρὰ ἡ ΕΗ τῆς ΕΖ buumlyuumlktuumlr dolayısıylaAnd [it] is equal ΕΗ to ΒΓ ἴση δὲ ἡ ΕΗ τῇ ΒΓ ΕΗ kenarı da ΕΖ kenarındangreater therefore is ΒΓ than ΕΖ μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ Ve eşittir ΕΗ ΒΓ kenarına

buumlyuumlktuumlr dolayısıyla ΒΓ ΕΖ kenarın-dan

CHAPTER ELEMENTS

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut angle τὴν δὲ γωνίαν ama accedilısıthan angle τῆς γωνίας accedilısındanhave greater μείζονα ἔχῃ buumlyuumlkse[namely] that by the equal sides τὴν ὑπὸ τῶν ἴσων εὐθειῶν [yani] eşit kenarlarcacontained περιεχομένην iccedilerilen(ler)also base καὶ τὴν βάσιν tabanı dathan base τῆς βάσεως tabanındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

ΒΓ

Δ

ΕΗ

Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birinebut base τὴν δὲ βάσιν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenler

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenleritwo sides ΑΒ and ΑΓ τὰς δύο πλευρὰς τὰς ΑΒ ΑΓ iki ΑΒ ve ΑΓ kenarıto two sides ΔΕ and ΔΖ ταῖς δύο πλευραῖς ταῖς ΔΕ ΔΖ iki ΔΕ ve ΔΖ kenarınahaving equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaand the base ΒΓ βάσις δὲ ἡ ΒΓ ve ΒΓ tabanıthan the base ΕΖ βάσεως τῆς ΕΖ ΕΖ tabanındanmdashlet it be greater μείζων ἔστω mdashbuumlyuumlk olsunI say that λέγω ὅτι İddia ediyorum kialso the angle ΒΑΓ καὶ γωνία ἡ ὑπὸ ΒΑΓ ΒΑΓ accedilısı dathan the angle ΕΔΖ γωνίας τῆς ὑπὸ ΕΔΖ ΕΔΖ accedilısındanis greater μείζων ἐστίν buumlyuumlktuumlr

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilse[it] is either equal to it or less ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων ya ona eşittir ya da ondan kuumlccediluumlkbut it is not equal ἴση μὲν οὖν οὐκ ἔστιν ama eşit değildirmdashΒΑΓ to ΕΔΖ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ mdashΒΑΓ ΕΔΖ accedilısınafor if it is equal ἴση γὰρ ἂν ἦν ccediluumlnkuuml eğer eşit isealso the base ΒΓ to ΕΖ καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ΒΓ tabanı da ΕΖ tabanına (eşittir)

but it is not οὐκ ἔστι δέ ama değilTherefore it is not equal οὐκ ἄρα ἴση ἐστὶ Dolayısıyla eşit değildirangle ΒΑΓ to ΕΔΖ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısınaneither is it less οὐδὲ μὴν ἐλάσσων ἐστὶν kuumlccediluumlk de değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısındanfor if it is less ἐλάσσων γὰρ ἂν ἦν ccediluumlnkuuml eğer kuumlccediluumlk isealso base ΒΓ than ΕΖ καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ ΒΓ tabanı da ΕΖ tabanından (kuumlccediluumlk-but it is not οὐκ ἔστι δέ tuumlr)therefore it is not less οὐκ ἄρα ἐλάσσων ἐστὶν ama değilΒΑΓ than angle ΕΔΖ ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ dolayısıyla kuumlccediluumlk değildirAnd it was shown that ἐδείχθη δέ ὅτι ΒΑΓ ΕΔΖ accedilısındanit is not equal οὐδὲ ἴση Ama goumlsterilmişti kitherefore it is greater μείζων ἄρα ἐστὶν eşit değildirΒΑΓ than ΕΔΖ ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ dolayısıyla buumlyuumlktuumlr

ΒΑΓ ΕΔΖ accedilısından

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo sides τὰς δύο πλευρὰς (birinin) iki kenarıto two sides δυσὶ πλευραῖς (diğerinin) iki kenarınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκάτερᾳ her biri birinebut base τὴν δὲ βασίν ama tabanıthan base τῆς βάσεως tabanındanhave greater μείζονα ἔχῃ buumlyuumlksealso angle καὶ τὴν γωνίαν accedilısı dathan angle τῆς γωνίας accedilısındanthey will have greater μείζονα ἕξει buumlyuumlk olacakmdashthat by the equal straights τὴν ὑπὸ τῶν ἴσων εὐθειῶν mdash(yani) eşit doğrularcacontained περιεχομένην iccedilerilenlermdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

If two triangles ᾿Εὰν δύο τρίγωνα Eğer iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarına

Let there be ῎Εστω Verilmiş olsuntwo triangles ΑΒΓ and ΔΕΖ δύο τρίγωνα τὰ ΑΒΓ ΔΕΖ iki ΑΒΓ ve ΔΕΖ uumlccedilgenithe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two angles ΔΕΖ and ΕΖΔ δυσὶ ταῖς ὑπὸ ΔΕΖ ΕΖΔ iki ΔΕΖ ve ΕΖΔ accedilılarına

CHAPTER ELEMENTS

having equal ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒΓ to ΔΕΖ τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ΑΒΓ ΔΕΖ accedilısınaand ΒΓΑ to ΕΖΔ τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ ve ΒΓΑ ΕΖΔ accedilısınaand let them also have ἐχέτω δὲ ayrıca olsunone side καὶ μίαν πλευρὰν bir kenarıto one side μιᾷ πλευρᾷ bir kenarınaequal ἴσην eşitfirst that near the equal angles πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις oumlnce esit accedilıların yanında olanΒΓ to ΕΖ τὴν ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarına

I say that λέγω ὅτι İddia ediyorum kithe remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarlarto the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarathey will have equal ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineΑΒ to ΔΕ τὴν μὲν ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaand ΑΓ to ΔΖ τὴν δὲ ΑΓ τῇ ΔΖ ve ΑΓ ΔΖ kenarınaalso the remaining angle καὶ τὴν λοιπὴν γωνίαν ayrıca kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaΒΑΓ to ΕΔΖ τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ΒΑΓ ΕΔΖ accedilısına

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Buumlyuumlk olanΑΒ ἡ ΑΒ ΑΒ olsunand let there be cut καὶ κείσθω ve kesilmiş olsunto ΔΕ equal τῇ ΔΕ ἴση ΔΕ kenarina eşitΒΗ ἡ ΒΗ ΒΗand suppose there has been joined καὶ ἐπεζεύχθω ve birleştirilmiş olsunΗΓ ἡ ΗΓ ΗΓ

Because then it is equal ᾿Επεὶ οὖν ἴση ἐστὶν Ccediluumlnkuuml o zaman eşittirΒΗ to ΔΕ ἡ μὲν ΒΗ τῇ ΔΕ ΒΗ ΔΕ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınathe two ΒΗ and ΒΓ δύο δὴ αἱ ΒΗ ΒΓ ΒΗ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand the angle ΗΒΓ καὶ γωνία ἡ ὑπὸ ΗΒΓ ve ΗΒΓ accedilısıto the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἴση ἐστίν eşittirtherefore the base ΗΓ βάσις ἄρα ἡ ΗΓ dolayısıyla ΗΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΗΒΓ καὶ τὸ ΗΒΓ τρίγωνον ve ΗΒΓ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılarawill be equal ἴσαι ἔσονται eşit olacaklarthose that the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν eşit kenarların karşıladıklarıEqual therefore is angle ΒΓΗ ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία Eşittir dolayısıyla ΒΓΗ accedilısıto ΔΖΕ τῇ ὑπὸ ΔΖΕ ΔΖΕ accedilısınaBut ΔΖΕ ἀλλὰ ἡ ὑπὸ ΔΖΕ Ama ΔΖΕto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais supposed equal ὑπόκειται ἴση eşit kabul edilmiştitherefore also ΒΓΗ καὶ ἡ ὑπὸ ΒΓΗ ἄρα dolayısıyla ΒΓΗ deto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἴση ἐστίν eşittirthe lesser to the greater ἡ ἐλάσσων τῇ μείζονι daha kuumlccediluumlk olan daha buumlyuumlk olanawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdır

Therefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla değildir eşit değilΑΒ to ΔΕ ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirIt is also the case that ἔστι δὲ καὶ Ayrıca durum şoumlyledirΒΓ to ΕΖ is equal ἡ ΒΓ τῇ ΕΖ ἴση ΒΓ ΕΖ kenarına eşittirthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ o zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso the angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒΓ accedilısı dato the angle ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΕΖ ΔΕΖ accedilısınais equal ἐστιν ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

But then again let them be Αλλὰ δὴ πάλιν ἔστωσαν Ama o zaman yine olsunlarmdash[those angles] equal sides αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ mdash kenarlar eşit [accedilıları]subtendingmdash ὑποτείνουσαι karşılayanmdashequal ἴσαι eşitas ΑΒ to ΔΕ ὡς ἡ ΑΒ τῇ ΔΕ ΑΒ ΔΕ kenarına gibiI say again that λέγω πάλιν ὅτι Yine iddia ediyorum kialso the remaining sides καὶ αἱ λοιπαὶ πλευραὶ kalan kenarlar dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarawill be equal ἴσαι ἔσονται eşit olacaklarΑΓ to ΔΖ ἡ μὲν ΑΓ τῇ ΔΖ ΑΓ ΔΖ kenarınaand ΒΓ to ΕΖ ἡ δὲ ΒΓ τῇ ΕΖ ve ΒΓ ΕΖ kenarınaand also the remaining angle ΒΑΓ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısı dato the remaining angle ΕΔΖ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση ἐστίν eşittir

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değil iseΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınaone of them is greater μία αὐτῶν μείζων ἐστίν biri daha buumlyuumlktuumlrLet be greater ἔστω μείζων Daha buumlyuumlk olsunif possible εἰ δυνατόν eğer muumlmkuumlnseΒΓ ἡ ΒΓ ΒΓand let there be cut καὶ κείσθω ve kesilmiş olsunto ΕΖ equal τῇ ΕΖ ἴση ΕΖ kenarına eşitΒΘ ἡ ΒΘ ΒΘand suppose there has been joined καὶ ἐπεζεύχθω ve kabul edilsin birleştirilmiş olduğuΑΘ ἡ ΑΘ ΑΘ kenarınınBecause also it is equal καὶ ἐπὲι ἴση ἐστὶν Ayrıca eşit olduğundanmdashΒΘ to ΕΖ ἡ μὲν ΒΘ τῇ ΕΖ mdashΒΘ ΕΖ kenarınaand ΑΒ to ΔΕ ἡ δὲ ΑΒ τῇ ΔΕ ve ΑΒ ΔΕ kenarınathen the two ΑΒ and ΒΘ δύο δὴ αἱ ΑΒ ΒΘ ΑΒ ve ΒΘ ikilisito the two ΔΕ and ΕΖ δυσὶ ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκαρέρᾳ her biri birineand they contain equal angles καὶ γωνίας ἴσας περιέχουσιν ama iccedilerirler eşit accedilılarıtherefore the base ΑΘ βάσις ἄρα ἡ ΑΘ dolayısıyla ΑΘ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand the triangle ΑΒΘ καὶ τὸ ΑΒΘ τρίγωνον ve ΑΒΘ uumlccedilgenito the triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

Fitzpatrick considers this way of denoting the line to be a lsquomis-takersquo apparently he thinks Euclid should (and perhaps did origi-nally) write ΗΒ for parallelism with ΔΕ But ΗΒ and ΒΗ are thesame line and for all we know Euclid preferred to write ΒΗ becauseit was in alphabetical order Netz [ Ch ] studies the general

Greek mathematical practice of using the letters in different orderfor the same mathematical object He concludes that changes in or-der are made on purpose though he does not address examples likethe present one

CHAPTER ELEMENTS

is equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve kalan accedilılarto the remaining angles ταῖς λοιπαῖς γωνίαις kalan accedilılaraare equal ἴσαι ἔσονται eşittirlerwhich the equal sides ὑφ᾿ ἃς αἱ ἴσας πλευραὶ eşit kenarlarınsubtend ὑποτείνουσιν karşıladıklarıTherefore equal is ἴση ἄρα ἐστὶν Dolayısıyla eşittirangle ΒΘΑ ἡ ὑπὸ ΒΘΑ γωνία ΒΘΑto ΕΖΔ τῇ ὑπὸ ΕΖΔ ΕΖΔ accedilısınaBut ΕΖΔ ἀλλὰ ἡ ὑπὸ ΕΖΔ Ama ΕΖΔto ΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınais equal ἐστιν ἴση eşittirthen of triangle ΑΘΓ τριγώνου δὴ τοῦ ΑΘΓ o zaman ΑΘΓ uumlccedilgenininthe exterior angle ΒΘΑ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ΒΘΑ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΒΓΑ τῇ ὑπὸ ΒΓΑ ΒΓΑ accedilısınawhich is impossible ὅπερ ἀδύνατον ki bu imkansızdırTherefore it is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΒΓ to ΕΖ ἡ ΒΓ τῇ ΕΖ ΒΓ ΕΖ kenarınatherefore it is equal ἴση ἄρα dolayısıyla eşittirAnd it is also ἐστὶ δὲ καὶ Ve yineΑΒ ἡ ΑΒ ΑΒto ΔΕ τῇ ΔΕ ΔΕ kenarınaequal ἴση eşittirThen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ O zaman ΑΒ ve ΒΓ ikilisito the two ΔΕ and ΕΖ δύο ταῖς ΔΕ ΕΖ ΔΕ ve ΕΖ ikilisineare equal ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birineand equal angles καὶ γωνίας ἴσας eşit accedilılarthey contain περιέχουσι iccedilerirlertherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ dolayısıyla ΑΓ tabanıto the base ΔΖ βάσει τῇ ΔΖ ΔΖ tabanınais equal ἴση ἐστίν eşittirand triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον ve ΑΒΓ uumlccedilgenito triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenineis equal ἴσον eşittirand the remaining angle ΒΑΓ καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ ve kalan ΒΑΓ accedilısıto the remaining angle ΕΔΖ τῇ λοιπῂ γωνίᾳ τῇ ὑπὸ ΕΔΖ kalan ΕΔΖ accedilısınais equal ἴση eşittir

If therefore two triangles ᾿Εὰν ἄρα δύο τρίγωνα Eğer dolayısıyla iki uumlccedilgenintwo angles τὰς δύο γωνίας iki accedilısıto two angles δυσὶ γωνίαις iki accedilısınahave equal ἴσας ἔχῃ eşitseeither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side καὶ μίαν πλευρὰν ve bir kenarto one side μιᾷ πλευρᾷ bir kenaraequal ἴσην eşitseeither that near the equal sides ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ya eşit accedilıların arasında olanor that subtending ἢ τὴν ὑποτείνουσαν ya da karşılayanone of the equal sides ὑπὸ μίαν τῶν ἴσων γωνιῶν eşit accedilılardan birinialso the remaining sides καὶ τὰς λοιπὰς πλευρὰς kalan kenarları dato the remaining sides ταῖς λοιπαῖς πλευραῖς kalan kenarlarınathey will have equal ἴσας ἕξει eşit olacakalso the remaining angle καὶ τὴν λοιπὴν γωνίαν kalan accedilıları dato the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilılarınamdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ

Δ

Ε Ζ

Η

Θ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilılarıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights Εἰς γὰρ δύο εὐθείας Ccediluumlnkuuml iki doğru uumlzerineΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ[suppose] the straight falling εὐθεῖα ἐμπίπτουσα [kabul edilsin] duumlşen[namely] ΕΖ ἡ ΕΖ ΕΖ doğrusununthe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΕΖ and ΕΖΔ τὰς ὑπὸ ΑΕΖ ΕΖΔ ΑΕΖ ve ΕΖΔ accedilılarınıequal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιείτω oluşturduğunu

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΒ to ΓΔ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ paraleldir ΑΒ ΓΔ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değilseextended ἐκβαλλόμεναι uzatılmışΑΒ and ΓΔ will meet αἱ ΑΒ ΓΔ συμπεσοῦνται ΑΒ ve ΓΔ buluşacaklareither in the ΒndashΔ parts ἤτοι ἐπὶ τὰ Β Δ μέρη ya ΒndashΔ parccedilalarındaor in the ΑndashΓ ἢ ἐπὶ τὰ Α Γ ya da ΑndashΓ parccedilalarındaSuppose they have been extended ἐκβεβλήσθωσαν Uzatılmış oldukları kabul edilsinand let them meet καὶ συμπιπτέτωσαν ve buluşşunlarin the ΒndashΔ parts ἐπὶ τὰ Β Δ μέρη ΒndashΔ parccedilalarındaat Η κατὰ τὸ Η Η noktasındaOf the triangle ΗΕΖ τριγώνου δὴ τοῦ ΗΕΖ ΗΕΖ uumlccedilgenininthe exterior angle ΑΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ΑΕΖ dış accedilısıis equal ἴση ἐστὶ eşittirto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtΕΖΗ τῇ ὑπὸ ΕΖΗ ΕΖΗ aşısınawhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore it is not [the case] that οὐκ ἄρα Dolayısıyla şoumlyle değildir (durum)ΑΒ and ΓΔ αἱ ΑΒ ΔΓ ΑΒ ve ΓΔextended ἐκβαλλόμεναι uzatılmışmeet in the ΒndashΔ parts συμπεσοῦνται ἐπὶ τὰ Β Δ μέρη buluşurlar ΒndashΔ parccedilalarındaSimilarly it will be shown that ὁμοίως δὴ δειχθήσεται ὅτι Benzer şekilde goumlsterilecek kineither on the ΑndashΓ οὐδὲ ἐπὶ τὰ Α Γ ΑndashΓ parccedilalarında daThose that in neither parts αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη Hiccedilbir parccediladameet συμπίπτουσαι buluşmayanlarare parallel παράλληλοί εἰσιν paraleldirtherefore parallel is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ dolayısıyla

paraleldir ΑΒ ΓΔ doğrusuna

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe alternate angles τὰς ἐναλλὰξ γωνίας ters accedilıları

CHAPTER ELEMENTS

equal to one another ἴσας ἀλλήλαις birbirine eşitmake ποιῇ yaparsaparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrularmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΓΔ

Η

Ε

Ζ

If on two straights ᾿Εὰν εἰς δύο εὐθείας Eğer iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται ἀλλήλαις birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular

For on the two straights ΑΒ and Εἰς γὰρ δύο εὐθείας τὰς ΑΒ ΓΔ Ccediluumlnkuuml ΑΒ ve ΓΔ doğruları uumlzerineΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ duumlşen ΕΖ doğrusu

the straight fallingmdashΕΖmdash τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ΕΗΒ dış accedilısınıthe exterior angle ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ iccedil ve karşıtto the interior and opposite angle τῇ ὑπὸ ΗΘΔ ΗΘΔ accedilısınaΗΘΔ ἴσην eşitequal ποιείτω mdashyaptığı varsayılsınmdashsuppose it makes ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanor the interior and in the same parts τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınınΒΗΘ and ΗΘΔ δυσὶν ὀρθαῖς iki dik accedilıyato two rights ἴσας eşit olduğuequal

I say that λέγω ὅτι İddia ediyorum kiparallel is παράλληλός ἐστιν paraleldirΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusuna

For since equal is ᾿Επεὶ γὰρ ἴση ἐστὶν Ccediluumlnkuuml eşit olduğundanΕΗΒ to ΗΘΔ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ ΕΗΒ ΗΘΔ accedilısınawhile ΕΗΒ to ΑΗΘ ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ aynı zamanda ΕΗΒ ΑΗΘ accedilısınais equal ἐστιν ἴση eşitkentherefore also ΑΗΘ to ΗΘΔ καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΑΗΘ de ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirand they are alternate καί εἰσιν ἐναλλάξ ve terstirlerparallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldirler dolayısıyla ΑΒ ve ΓΔ

Alternatively since ΒΗΘ and ΗΘΔ Πάλιν ἐπεὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ Ya da ΒΗΘ ve ΗΘΔto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirrand also are ΑΗΘ and ΒΗΘ εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ ΒΗΘ ve ΑΗΘ ve ΒΗΘ deto two rights δυσὶν ὀρθαῖς iki dik accedilıya

It is perhaps impossible to maintain the Greek word order com-prehensibly in English The normal English order would be lsquoIf astraight line falling on two straight linesrsquo But the proposition is

ultimately about the two straight lines perhaps that is why Euclidmentions them before the one straight line that falls on them

equal ἴσαι eşittirtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınaare equal ἴσαι εἰσίν eşittirlesuppose the common has been taken κοινὴ ἀφῃρήσθω varsayılsın ccedilıkartılmış olduğu ortak

away olanmdashΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘ accedilısınıntherefore the remaining ΑΗΘ λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ dolayısıyla ΑΗΘ kalanıto the remaining ΗΘΔ λοιπῇ τῇ ὑπὸ ΗΘΔ ΗΘΔ kalanınais equal ἐστιν ἴση eşittiralso they are alternate καί εἰσιν ἐναλλάξ ve bunlar terstirlerparallel therefore are ΑΒ and ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ paraleldir dolayısıyla ΑΒ ve ΓΔ

If therefore on two straights ᾿Εὰν ἄρα εἰς δύο εὐθείας Eğer dolayısıyla iki doğru uumlzerinea straight falling εὐθεῖα ἐμπίπτουσα duumlşen bir doğruthe exterior angle τὴν ἐκτὸς γωνίαν dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta kalan accedilıyamake equal ἴσην ποιῇ eşit yaparsaor the interior and in the same parts ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη veya iccedil ve aynı tarafta kalanlarıto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal ἴσας eşitparallel will be to one another παράλληλοι ἔσονται birbirine paralel olacakthe straights αἱ εὐθεῖαι doğrular mdashgoumlsterilmesi gereken tammdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

The straight falling on parallel ῾Η εἰς τὰς παραλλήλους εὐθείας εὐθεῖα Paralel doğrular uumlzerine duumlşen birstraights ἐμπίπτουσα doğru

the alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ birbirine eşit yaparand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıt accedilıyaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı tarafta kalanlarıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

For on the parallel straights Εἰς γὰρ παραλλήλους εὐθείας Ccediluumlnkuuml paralelΑΒ and ΓΔ τὰς ΑΒ ΓΔ ΑΒ ve ΓΔ doğruları uumlzerinelet the straight ΕΖ fall εὐθεῖα ἐμπιπτέτω ἡ ΕΖ ΕΖ doğrusu duumlşsuumln

I say that λέγω ὅτι İddia ediyorum kithe alternate angles τὰς ἐναλλὰξ γωνίας tersΑΗΘ and ΗΘΔ τὰς ὑπὸ ΑΗΘ ΗΘΔ ΑΗΘ ve ΗΘΔ accedilılarınıequal ἴσας eşitit makes ποιεῖ yaparand the exterior angle ΕΗΒ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ ve ΕΗΒ dış accedilısınıto the interior and opposite ΗΘΔ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ iccedil ve karşıt ΗΘΔ accedilısınaequal ἴσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakiΒΗΘ and ΗΘΔ τὰς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ile ΗΘΔ accedilılarınıto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşit

CHAPTER ELEMENTS

For if it is unequal Εἰ γὰρ ἄνισός ἐστιν Ccediluumlnkuuml eğer eşit değilseΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaone of them is greater μία αὐτῶν μείζων ἐστίν biri buumlyuumlktuumlrLet the greater be ΑΗΘ ἔστω μείζων ἡ ὑπὸ ΑΗΘ Buumlyuumlk olan ΑΗΘ olsunlet be added in common κοινὴ προσκείσθω eklenmiş olsun her ikisine deΒΗΘ ἡ ὑπὸ ΒΗΘ ΒΗΘtherefore ΑΗΘ and ΒΗΘ αἱ ἄρα ὑπὸ ΑΗΘ ΒΗΘ dolayısıyla ΑΗΘ ve ΒΗΘthan ΒΗΘ and ΗΘΔ τῶν ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarındanare greater μείζονές εἰσιν buumlyuumlktuumlrlerHowever ΑΗΘ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΑΗΘ ΒΗΘ Fakat ΑΗΘ ve ΒΗΘto two rights δυσὶν ὀρθαῖς iki dik accedilıyaequal are ἴσαι εἰσίν eşittirlerTherefore [also] ΒΗΘ and ΗΘΔ [καὶ] αἱ ἄρα ὑπὸ ΒΗΘ ΗΘΔ Dolayısıyla ΒΗΘ ve ΗΘΔ [da]than two rights δύο ὀρθῶν iki dik accedilıdanless are ἐλάσσονές εἰσιν kuumlccediluumlktuumlrlerAnd [straights] from [angles] that αἱ δὲ ἀπ᾿ ἐλασσόνων Ve kuumlccediluumlk olanlardan

are less ἢ δύο ὀρθῶν iki dik accedilıdanthan two rights ἐκβαλλόμεναι sonsuza uzatılanlar [doğrular]extended to the infinite εἰς ἄπειρον birbirinin uumlzerine duumlşerlerfall together συμπίπτουσιν Dolayısıyla ΑΒ ve ΓΔTherefore ΑΒ and ΓΔ αἱ ἄρα ΑΒ ΓΔ uzatılınca sonsuzaextended to the infinite ἐκβαλλόμεναι εἰς ἄπειρον birbirinin uumlzerine duumlşeceklerwill fall together συμπεσοῦνται Ama onlar birbirinin uumlzerine duumlşme-But they do not fall together οὐ συμπίπτουσι δὲ zlerby their being assumed parallel διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι paralel oldukları kabul edildiğindenTherefore is not unequal οὐκ ἄρα ἄνισός ἐστιν Dolayısıyla eşit değil değildirΑΗΘ to ΗΘΔ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ ΑΗΘ ΗΘΔ accedilısınaTherefore it is equal ἴση ἄρα Dolayısıyla eşittirHowever ΑΗΘ to ΕΗΒ ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ Ancak ΑΗΘ ΕΗΒ accedilısınais equal ἐστιν ἴση eşittirtherefore also ΕΗΒ to ΗΘΔ καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ dolayısıyla ΕΗΒ da ΗΘΔ accedilısınais equal ἐστιν ἴση eşittirlet ΒΗΘ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ eklenmiş olsun her ikisine de ΒΗΘtherefore ΕΗΒ and ΒΗΘ αἱ ἄρα ὑπὸ ΕΗΒ ΒΗΘ dolayısıyla ΕΗΒ ve ΒΗΘto ΒΗΘ and ΗΘΔ ταῖς ὑπὸ ΒΗΘ ΗΘΔ ΒΗΘ ve ΗΘΔ accedilılarınais equal ἴσαι εἰσίν eşittirBut ΕΗΒ and ΒΗΘ ἀλλὰ αἱ ὑπὸ ΕΗΒ ΒΗΘ Ama ΕΗΒ ve ΒΗΘto two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirlerTherefore also ΒΗΘ and ΗΘΔ καὶ αἱ ὑπὸ ΒΗΘ ΗΘΔ ἄρα Dolayısıyla ΒΗΘ ve ΗΘΔ dato two rights δύο ὀρθαῖς iki dik accedilıyaare equal ἴσαι εἰσίν eşittirler

Therefore the on-parallel-straights ῾Η ἄρα εἰς τὰς παραλλήλους εὐθείας εὐ- Dolayısıyla paralel doğrular uumlzerinestraight θεῖα doğru

falling ἐμπίπτουσα duumlşerkenthe alternate angles τάς τε ἐναλλὰξ γωνίας ters accedilılarımakes equal to one another ἴσας ἀλλήλαις ποιεῖ eşit yapar birbirineand the exterior καὶ τὴν ἐκτὸς ve dış accedilıyıto the interior and opposite τῇ ἐντὸς καὶ ἀπεναντίον iccedil ve karşıtaequal ίσην eşitand the interior and in the same parts καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη ve iccedil ve aynı taraftakileri sto two rights equal δυσὶν ὀρθαῖς ἴσας iki dik accedilıya eşitmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ Δ

Ε

Ζ

Η

Θ

[Straights] to the same straight Αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι Aynı doğruyaparallel καὶ ἀλλήλαις εἰσὶ παράλληλοι paralel doğrularalso to one another are parallel birbirlerine de paraleldir

Let be ῎Εστω Olsuneither of ΑΒ and ΓΔ ἑκατέρα τῶν ΑΒ ΓΔ ΑΒ ve ΓΔ doğrularının her birito ΓΔ parallel τῇ ΕΖ παράλληλος ΓΔ doğrusuna paralel

I say that λέγω ὅτι İddia ediyorum kialso ΑΒ to ΓΔ is parallel καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος ΑΒ da ΓΔ doğrusuna paraleldir

For let fall on them a straight ΗΚ ᾿Εμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ Ccediluumlnkuuml uumlzerlerine bir ΗΚ doğrusuduumlşmuumlş olsun

Then since on the parallel straights Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας O zaman paralelΑΒ and ΕΖ τὰς ΑΒ ΕΖ ΑΒ ve ΕΖ doğrularının uumlzerinea straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ bir doğru duumlşmuumlş olduğundan [yani]equal therefore is ΑΗΚ to ΗΘΖ ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ΗΚMoreover πάλιν eşittir dolayısıyla ΑΗΚ ΗΘΖ accedilısınasince on the parallel straights ἐπεὶ εἰς παραλλήλους εὐθείας DahasıΕΖ and ΓΔ τὰς ΕΖ ΓΔ paralela straight has fallen [namely] ΗΚ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ ΕΖ ve ΓΔ doğrularının uumlzerineequal is ΗΘΖ to ΗΚΔ ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ bir doğru duumlşmuumlş olduğundan [yani]And it was shown also that ἐδείχθη δὲ καὶ ΗΚΑΗΚ to ΗΘΖ is equal ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση eşittir ΗΘΖ ΗΚΔ accedilısınaAlso ΑΗΚ therefore to ΗΚΔ καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ Ve goumlsterilmişti kiis equal ἐστιν ἴση ΑΗΚ ΗΘΖ accedilısına eşittirand they are alternate καί εἰσιν ἐναλλάξ Ve ΑΗΚ dolayısıyla ΗΚΔ accedilısınaParallel therefore is ΑΒ to ΓΔ παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ eşittir

ve bunlar terstirlerParaleldir dolayısıyla ΑΒ ΓΔ

doğrusuna

Therefore [straights] [Αἱ ἄρα Dolayısıylato the same straight τῇ αὐτῇ εὐθείᾳ aynı doğruya paralellerparallel παράλληλοι birbirlerine de paraleldiralso to one another are parallel καὶ ἀλλήλαις εἰσὶ παράλληλοι] mdashgoumlsterilmesi gereken tam buydumdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι

Α Β

Γ Δ

Ε Ζ

Η

Θ

Κ

Through the given point Διὰ τοῦ δοθέντος σημείου Verilen bir noktadanto the given straight parallel τῇ δοθείσῃ εὐθείᾳ παράλληλον verilen bir doğruya paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Let be ῎Εστω Olsunthe given point Α τὸ μὲν δοθὲν σημεῖον τὸ Α verilen nokta Αand the given straight ΒΓ ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ ve verilen doğru ΒΓ

It is necessary then δεῖ δὴ Şimdi gereklidir

CHAPTER ELEMENTS

through the point Α διὰ τοῦ Α σημείου Α noktasındanto the straight ΒΓ parallel τῇ ΒΓ εὐθείᾳ παράλληλον ΒΓ doğrusuna paralela straight line to draw εὐθεῖαν γραμμὴν ἀγαγεῖν bir doğru ccedilizmek

Suppose there has been chosen Εἰλήφθω Varsayılsın seccedililmiş olduğuon ΒΓ ἐπὶ τῆς ΒΓ ΒΓ uumlzerindea random point Δ τυχὸν σημεῖον τὸ Δ rastgele bir Δ noktasınınand there has been joined ΑΔ καὶ ἐπεζεύχθω ἡ ΑΔ ve ΑΔ doğrusunun birleştirilmişand there has been constructed καὶ συνεστάτω olduğuon the straight ΔΑ πρὸς τῇ ΔΑ εὐθείᾳ ve inşa edilmiş olduğuand at the point Α of it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α ΔΑ doğrusundato the angle ΑΔΓ equal τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ve onun Α noktasındaΔΑΕ ἡ ὑπὸ ΔΑΕ ΑΔΓ accedilısına eşitand suppose there has been extended καὶ ἐκβεβλήσθω ΔΑΕ accedilısınınin straights with ΕΑ ἐπ᾿ εὐθείας τῇ ΕΑ ve kabul edilsin uzatılmış olsunthe straight ΑΖ εὐθεῖα ἡ ΑΖ ΕΑ ile aynı doğruda

ΑΖ doğrusu

And because Καὶ ἐπεὶ Ve ccediluumlnkuumlon the two straights ΒΓ and ΕΖ εἰς δύο εὐθείας τὰς ΒΓ ΕΖ ΒΓ ve ΕΖ doğruları uumlzerinethe straight line falling ΑΔ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ duumlşerken ΑΔ doğrusuthe alternate angles τὰς ἐναλλὰξ γωνίας tersΕΑΔ and ΑΔΓ τὰς ὑπὸ ΕΑΔ ΑΔΓ ΕΑΔ ve ΑΔΓ accedilılarınıequal to one another has made ἴσας ἀλλήλαις πεποίηκεν eşit yapmıştır birbirineparallel therefore is ΕΑΖ to ΒΓ παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ paraleldir dolayısıyla ΕΑΖ ΒΓ

doğrusuna

Therefore through the given point Α Διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α Dolayısıyla verilen Α noktasındanto the given straight ΒΓ parallel τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος verilen ΒΓ doğrusuna paralela straight line has been drawn ΕΑΖ εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ bir doğru ΕΑΖ ccedilizilmiş oldumdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Α

Β ΓΔ

Ε Ζ

Of any triangle Παντὸς τριγώνου Herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Let there be ῎Εστω Verilmiş olsunthe triangle ΑΒΓ τρίγωνον τὸ ΑΒΓ ΑΒΓ uumlccedilgeniand suppose there has been extended καὶ προσεκβεβλήσθω ve varsayılsın uzatılmış olduğuits one side ΒΓ to Δ αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ bir ΒΓ kenarının Δ noktasına

I say that λέγω ὅτι İddia ediyorum kithe exterior angle ΑΓ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ ΑΓΔ dış accedilısı eşittir

to the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki iccedil ve karşıtΓΑΒ and ΑΒΓ ταῖς ὑπὸ ΓΑΒ ΑΒΓ ΓΑΒ ve ΑΒΓ accedilısınaand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıΑΒΓ ΒΓΑ and ΓΑΒ αἱ ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

For suppose there has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml varsayılsın ccedilizilmiş olduğuthrough the point Γ διὰ τοῦ Γ σημείου Γ noktasındanto the straight ΑΒ parallel τῇ ΑΒ εὐθείᾳ παράλληλος ΑΒ doğrusuna paralelΓΕ ἡ ΓΕ ΓΕ doğrusunun

And since parallel is ΑΒ to ΓΕ Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ Ve paralel olduğundan ΑΒ ΓΕand on these has fallen ΑΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ doğrusunathe alternate angles ΒΑΓ and ΑΓΕ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ ΑΓΕ ve bunların uumlzerine duumlştuumlğuumlnden ΑΓequal to one another are ἴσαι ἀλλήλαις εἰσίν ters ΒΑΓ ve ΑΓΕ accedilılarıMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν eşittirler birbirlerineΑΒ to ΓΕ ἡ ΑΒ τῇ ΓΕ Dahası paralel olduğundanand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ΑΒ ΓΕ doğrusunathe straight ΒΔ εὐθεῖα ἡ ΒΔ and bunların uumlzerine duumlştuumlğuumlndenthe exterior angle ΕΓΔ is equal ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ ΒΔ doğrusuto the interior and opposite ΑΒΓ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ ΕΓΔ dış accedilısı eşittirAnd it was shown that ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ iccedil ve karşıt ΑΒΓ accedilısınaalso ΑΓΕ to ΒΑΓ [is] equal ἴση Ve goumlsterilmişti kiTherefore the whole angle ΑΓΔ ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ΑΓΕ da ΒΑΓ accedilısına eşittiris equal ἴση ἐστὶ Dolayısıyla accedilının tamamı ΑΓΔto the two interior and opposite angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον eşittirΒΑΓ and ΑΒΓ ταῖς ὑπὸ ΒΑΓ ΑΒΓ iccedil ve karşıt

ΒΑΓ ve ΑΒΓ accedilılarına

Let be added in common ΑΓΒ Κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ Eklenmiş olsun ΑΓΒ ortak olarakTherefore ΑΓΔ and ΑΓΒ αἱ ἄρα ὑπὸ ΑΓΔ ΑΓΒ Dolayısıyla ΑΓΔ ve ΑΓΒ accedilılarıto the three ΑΒΓ ΒΓΑ and ΓΑΒ τρισὶ ταῖς ὑπὸ ΑΒΓ ΒΓΑ ΓΑΒ ΑΒΓ ΒΓΑ ve ΓΑΒ uumlccedilluumlsuumlneequal are ἴσαι εἰσίν eşittirHowever ΑΓΔ and ΑΓΒ ἀλλ᾿ αἱ ὑπὸ ΑΓΔ ΑΓΒ Fakat ΑΓΔ ve ΑΓΒ accedilılarıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittiralso ΑΓΒ ΓΒΑ and ΓΑΒ therefore καὶ αἱ ὑπὸ ΑΓΒ ΓΒΑ ΓΑΒ ἄρα ΑΓΒ ΓΒΑ ve ΓΑΒ da dolayısıylato two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittir

Therefore of any triangle Παντὸς ἄρα τριγώνου Dolayısıyla herhangi bir uumlccedilgeninone of the sides being extended μιᾶς τῶν πλευρῶν προσεκβληθείσης kenarlarından biri uzatıldığındathe exterior angle ἡ ἐκτὸς γωνία dış accedilıto the two opposite interior angles δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον iki karşıt iccedil accedilıyais equal ἴση ἐστίν eşittirand the trianglersquos three interior angles καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι ve uumlccedilgenin uumlccedil iccedil accedilısıto two rights equal are δυσὶν ὀρθαῖς ἴσαι εἰσίν iki dik accedilıya eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α

Β Γ Δ

Ε

Straights joining equals and paral- Αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ Eşit ve paralellerin aynı taraflarınılels to the same parts αὐτὰ μέρη ἐπιζευγνύουσαι εὐ- birleştiren doğruların

also themselves equal and parallel are θεῖαι kendileri de eşit ve paraleldirlerκαὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν

CHAPTER ELEMENTS

Let be ῎Εστωσαν Olsunequals and parallels ἴσαι τε καὶ παράλληλοι eşit ve paralellerΑΒ and ΓΔ αἱ ΑΒ ΓΔ ΑΒ ve ΓΔand let join these καὶ ἐπιζευγνύτωσαν αὐτὰς ve bunların birleştirsinin the same parts ἐπὶ τὰ αὐτὰ μέρη aynı taraflarınıstraights ΑΓ and ΒΔ εὐθεῖαι αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ doğruları

I say that λέγω ὅτι İddia ediyorum kialso ΑΓ and ΒΔ καὶ αἱ ΑΓ ΒΔ ΑΓ ve ΒΔ daequal and parallel are ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirler

Suppose there has been joined ΒΓ ᾿Επεζεύχθω ἡ ΒΓ Varsayılsın birleştirilmiş olduğu ΒΓAnd since parallel is ΑΒ to ΓΔ καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ doğrusununand on these has fallen ΒΓ καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ Ve paralel olduğundan ΑΒ ΓΔthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ doğrusunaequal to one another are ἴσαι ἀλλήλαις εἰσίν ve bunların uumlzerine duumlştuumlğuumlnden ΒΓAnd since equal is ΑΒ to ΓΔ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıand common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ birbirlerine eşittirlerthen the two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ Ve eşit olduğundan ΑΒ ΓΔto the two ΒΓ and ΓΔ δύο ταῖς ΒΓ ΓΔ doğrusunaequal are ἴσαι εἰσίν ve ΒΓ ortakalso angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ ΑΒ ve ΒΓ ikilisito angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ΒΓ ve ΓΔ ikilisine[is] equal ἴση eşittirtherefore the base ΑΓ βάσις ἄρα ἡ ΑΓ ΑΒΓ accedilısı dato the base ΒΔ βάσει τῇ ΒΔ ΒΓΔ accedilısınais equal ἐστιν ἴση eşittirand the triangle ΑΒΓ καὶ τὸ ΑΒΓ τρίγωνον dolayısıyla ΑΓ tabanıto the triangle ΒΓΔ τῷ ΒΓΔ τριγώνῳ ΒΔ tabanınais equal ἴσον ἐστίν eşittirand the remaining angles καὶ αἱ λοιπαὶ γωνίαι ve ΑΒΓ uumlccedilgenito the remaining angles ταῖς λοιπαῖς γωνίαις ΒΓΔ uumlccedilgenineequal will be ἴσαι ἔσονται eşittireither to either ἑκατέρα ἑκατέρᾳ ve kalan accedilılarwhich the equal sides subtend ὑφ᾿ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν kalan accedilılaraequal therefore ἴση ἄρα eşit olacaklarthe ΑΓΒ angle to ΓΒΔ ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ her biri birineAnd since on the two straights καὶ ἐπεὶ εἰς δύο εὐθείας eşit kenarları goumlrenlerΑΓ and ΒΔ τὰς ΑΓ ΒΔ eşittir dolayısıylathe straight fallingmdashΒΓmdash εὐθεῖα ἐμπίπτουσα ἡ ΒΓ ΑΓΒ ΓΒΔ accedilısınaalternate angles equal to one another τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις Ve uumlzerine ikihas made πεποίηκεν ΑΓ ve ΒΔ doğrularınınparallel therefore is ΑΓ to ΒΔ παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ duumlşen doğrumdashΒΓmdashAnd it was shown to it also equal ἐδείχθη δὲ αὐτῇ καὶ ἴση birbirine eşit ters accedilılar

yapmıştırparaleldir dolayısıyla ΑΓ ΒΔ

doğrusunaVe eşit olduğu da goumlsterilmişti

Therefore straights joining equals Αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐ- Dolayısıyla eşit ve paralellerin aynıand parallels to the same parts πὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι taraflarını birleştiren doğru-

also themselves equal and parallel are εὐθεῖαι larınmdashjust what it was necessary to καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν kendileri de eşitshow ὅπερ ἔδει δεῖξαι ve paraleldirler mdashgoumlsterilmesi

gereken tam buydu

Β Α

Δ Γ

Of parallelogram areas Τῶν παραλληλογράμμων χωρίων Paralelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıare equal to one another ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the diameter cuts them in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει ve koumlşegen onları ikiye boumller

Let there be ῎Εστω Verilmiş olsuna parallelogram area παραλληλόγραμμον χωρίον bir paralelkenar alanΑΓΔΒ τὸ ΑΓΔΒ ΑΓΔΒa diameter of it ΒΓ διάμετρος δὲ αὐτοῦ ἡ ΒΓ ve onun bir koumlşegeni ΒΓ

I say that λέγω ὅτι iddia ediyorum kiof the ΑΓΔΒ parallelogram τοῦ ΑΓΔΒ παραλληλογράμμου ΑΓΔΒ paralelkenarınınthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirineand the ΒΓ diameter it cuts in two καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει ve ΒΓ koumlşegeni onu ikiye boumller

For since parallel is ᾿Επεὶ γὰρ παράλληλός ἐστιν Ccediluumlnkuuml paralel olduğundanΑΒ to ΓΔ ἡ ΑΒ τῇ ΓΔ ΑΒ ΓΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndena straight ΒΓ εὐθεῖα ἡ ΒΓ bir ΒΓ doğrusuthe alternate angles ΑΒΓ and ΒΓΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ ΒΓΔ ters ΑΒΓ ve ΒΓΔ accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineMoreover since parallel is πάλιν ἐπεὶ παράλληλός ἐστιν Dahası paralel olduğundanΑΓ to ΒΔ ἡ ΑΓ τῇ ΒΔ ΑΓ ΒΔ doğrusunaand on these has fallen καὶ εἰς αὐτὰς ἐμπέπτωκεν ve bunların uumlzerine duumlştuumlğuumlndenΒΓ ἡ ΒΓ ΒΓthe alternate angles ΑΓΒ and ΓΒΔ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ ΓΒΔ ters accedilılar ΑΓΒ ve ΓΒΔequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerineThen two triangles there are δύο δὴ τρίγωνά ἐστι Şimdi iki uumlccedilgen vardırΑΒΓ and ΒΓΔ τὰ ΑΒΓ ΒΓΔ ΑΒΓ ve ΒΓΔthe two angles ΑΒΓ and ΒΓΑ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ ΒΓΑ iki ΑΒΓ ve ΒΓΑ accedilılarıto the two ΒΓΔ and ΓΒΔ δυσὶ ταῖς ὑπὸ ΒΓΔ ΓΒΔ iki ΒΓΔ ve ΓΒΔ accedilılarınaequal having ἴσας ἔχοντα eşit olaneither to either ἑκατέραν ἑκατέρᾳ her biri birineand one side to one side equal καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ve bir kenarı bir kenarına eşit olanthat near the equal angles τὴν πρὸς ταῖς ἴσαις γωνίαις eşit accedilıların yanında olantheir common ΒΓ κοινὴν αὐτῶν τὴν ΒΓ onların ortak ΒΓ kenarıalso then the remaining sides καὶ τὰς λοιπὰς ἄρα πλευρὰς o zaman kalan kenarları dato the remaining sides ταῖς λοιπαῖς kalan kenarlarınaequal they will have ἴσας ἕξει eşit olacaklareither to either ἑκατέραν ἑκατέρᾳ her biri birineand the remaining angle καὶ τὴν λοιπὴν γωνίαν ve kalan accedilıto the remaining angle τῇ λοιπῇ γωνίᾳ kalan accedilıyaequal therefore ἴση ἄρα eşit dolayısıylathe ΑΒ side to ΓΔ ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ ΑΒ kenarı ΓΔ kenarınaand ΑΓ to ΒΔ ἡ δὲ ΑΓ τῇ ΒΔ ve ΑΓ ΒΔ kenarınaand yet equal is the ΒΑΓ angle καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία ve eşittir ΒΑΓ accedilısıto ΓΔΒ τῇ ὑπὸ ΓΔΒ ΓΔΒ accedilısınaAnd since equal is the ΑΒΓ angle καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία Ve eşit olduğundan ΑΒΓto ΒΓΔ τῇ ὑπὸ ΒΓΔ ΒΓΔ accedilısınaand ΓΒΔ to ΑΓΒ ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ ve ΓΒΔ ΑΓΒ accedilısınatherefore the whole ΑΒΔ ὅλη ἄρα ἡ ὑπὸ ΑΒΔ dolayısıyla accedilının tamamı ΑΒΔto the whole ΑΓΔ ὅλῃ τῇ ὑπὸ ΑΓΔ accedilının tamamına ΑΓΔis equal ἐστιν ἴση eşittirAnd was shown also ἐδείχθη δὲ καὶ Ve goumlsterilmişti ayrıcaΒΑΓ to ΓΔΒ equal ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση ΒΑΓ ile ΓΔΒ accedilısının eşitliği

Therefore of parallelogram areas Τῶν ἄρα παραλληλογράμμων χωρίων Dolayısıylaparalelkenar alanlarınopposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι karşıt kenar ve accedilılarıequal to one another are ἴσαι ἀλλήλαις εἰσίν eşittir birbirlerine

CHAPTER ELEMENTS

I say then that Λέγω δή ὅτι Şimdi iddia ediyorum kialso the diameter them cuts in two καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει koumlşegen de onları ikiye keser

For since equal is ΑΒ to ΓΔ ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ Ccediluumlnkuuml eşit olduğundan ΑΒ ΓΔ ke-and common [is] ΒΓ κοινὴ δὲ ἡ ΒΓ narınathe two ΑΒ and ΒΓ δύο δὴ αἱ ΑΒ ΒΓ ve ΒΓ ortakto the two ΓΔ and ΒΓ δυσὶ ταῖς ΓΔ ΒΓ ΑΒ ve ΒΓ ikilisiequal are ἴσαι εἰσὶν - ΓΔ ve ΒΓ ikilisineeither to either ἑκατέρα ἑκατέρᾳ eşittirlerand angle ΑΒΓ καὶ γωνία ἡ ὑπὸ ΑΒΓ her biri birineto angle ΒΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ve ΑΒΓ accedilısıequal ἴση ΒΓΔ accedilısınaTherefore also the base ΑΓ καὶ βάσις ἄρα ἡ ΑΓ eşittirto the base ΔΒ τῇ ΔΒ Dolayısıyla ΑΓ tabanı daequal ἴση ΔΒ tabanınaTherefore also the ΑΒΓ triangle καὶ τὸ ΑΒΓ [ἄρα] τρίγωνον eşittirto the ΒΓΔ triangle τῷ ΒΓΔ τριγώνῳ Dolayısıyla ΑΒΓ uumlccedilgeni deis equal ἴσον ἐστίν ΒΓΔ uumlccedilgenine

eşittir

Therefore the ΒΓ diameter cuts in two ῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει Dolayısıyla ΒΓ koumlşegeni ikiye boumllerthe ΑΒΓΔ parallelogram τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarınımdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

ΔΓ

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittir

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΒΓΔ τὰ ΑΒΓΔ ΕΒΓΖ ΑΒΓΔ ve ΕΒΓΔon the same base ΓΒ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΓΒ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΖ and ΒΓ ταῖς ΑΖ ΒΓ ΑΖ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔto the parallelogram ΕΒΓΖ τῷ ΕΒΓΖ παραλληλογράμμῳ ΕΒΓΖ paralelkenarına

For since ᾿Επεὶ γὰρ Ccediluumlnkuumla parallelogram is ΑΒΓΔ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ bir paralelkenar olduğundan ΑΒΓΔequal is ΑΔ to ΒΓ ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ eşittir ΑΔ ΒΓ kenarınaSimilarly then also διὰ τὰ αὐτὰ δὴ καὶ Benzer şekilde o zamanΕΖ to ΒΓ is equal ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση ΕΖ ΒΓ kenarına eşittirso that also ΑΔ to ΕΖ is equal ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση boumlylece ΑΔ da ΕΖ kenarına eşittirand common [is] ΔΕ καὶ κοινὴ ἡ ΔΕ ve ortaktır ΔΕtherefore ΑΕ as a whole ὅλη ἄρα ἡ ΑΕ dolayısıyla ΑΕ bir buumltuumln olarakto ΔΖ as a whole ὅλῃ τῇ ΔΖ ΔΖ kenarınais equal ἐστιν ἴση eşittir

Is also ΑΒ to ΔΓ equal ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση ΑΒ da ΔΓ kenarına eşittirThen the two ΕΑ and ΑΒ δύο δὴ αἱ ΕΑ ΑΒ O zaman ΕΑ ve ΑΒ ikilisito the two ΖΔ and ΔΓ δύο ταῖς ΖΔ ΔΓ ΖΔ ve ΔΓ ikilisineequal are ἴσαι εἰσὶν eşittirlereither to either ἑκατέρα ἑκατέρᾳ her biri birinealso angle ΖΔΓ καὶ γωνία ἡ ὑπὸ ΖΔΓ ve ΖΔΓ accedilısı dato ΕΑΒ γωνίᾳ τῇ ὑπὸ ΕΑΒ ΕΑΒ accedilısınais equal ἐστιν ἴση eşittirthe exterior to the interior ἡ ἐκτὸς τῇ ἐντός dış accedilı iccedil accedilıyatherefore the base ΕΒ βάσις ἄρα ἡ ΕΒ dolayısıyla ΕΒ tabanıto the base ΖΓ βάσει τῇ ΖΓ ΖΓ tabanınais equal ἴση ἐστίν eşittirand triangle ΕΑΒ καὶ τὸ ΕΑΒ τρίγωνον ve ΕΑΒ uumlccedilgenito triangle ΔΖΓ τῷ ΔΖΓ τριγώνῳ ΔΖΓ uumlccedilgenineequal will be ἴσον ἔσται eşit olacaksuppose has been removed commonly κοινὸν ἀφῃρήσθω τὸ ΔΗΕ kaldırılmış olsun ortak olarakΔΗΕ λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον ΔΗΕtherefore the trapezium ΑΒΗΔ that λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ dolayısıyla kalan ΑΒΗΔ yamuğu

remains ἐστὶν ἴσον kalan ΕΗΓΖ yamuğunato the trapezium ΕΗΓΖ that remains κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον eşittiris equal ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον eklenmiş olsun her ikisine birdenlet be added in common ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ΗΒΓ uumlccedilgenithe triangle ΗΒΓ ἴσον ἐστίν dolayısıyla ΑΒΓΔ paralelkenarınıntherefore the parallelogram ΑΒΓΔ as tamamı

a whole ΕΒΓΖ paralelkenarının tamamınato the parallelogram ΕΒΓΖ as a whole eşittiris equal

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısısyla paralelkenarlaron the same base being τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabanda olanand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarequal to one another are ἴσα ἀλλήλοις ἐστίν birbirlerine eşittirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

ΑΔ

Γ

ΕΖ

Η

Parallelograms Τὰ παραλληλόγραμμα Paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerine

Let there be ῎Εστω Verilmiş olsunparallelograms παραλληλόγραμμα paralelkenarlarΑΒΓΔ and ΕΖΗΘ τὰ ΑΒΓΔ ΕΖΗΘ ΑΒΓΔ ve ΕΖΗΘon equal bases ἐπὶ ἴσων βάσεων ὄντα eşitΒΓ and ΖΗ τῶν ΒΓ ΖΗ ΒΓ ve ΖΗ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΘ and ΒΗ ταῖς ΑΘ ΒΗ ΑΘ ve ΒΗ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirparallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔto ΕΖΗΘ τῷ ΕΖΗΘ ΕΖΗΘ paralelkenarına

For suppose have been joined ᾿Επεζεύχθωσαν γὰρ Ccediluumlnkuuml varsayılsın birleştirilmiş

CHAPTER ELEMENTS

ΒΕ and ΓΘ αἱ ΒΕ ΓΘ olduğuΒΕ ile ΓΘ kenarlarının

And since equal are ΒΓ and ΖΗ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ Ve eşit olduğundan ΒΓ ile ΖΗbut ΖΗ to ΕΘ is equal ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση ama ΖΗ ΕΘ kenarına eşittirtherefore also ΒΓ to ΕΘ is equal καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση dolayısıyla ΒΓ da ΕΘ kenarına eşittirAnd [they] are also parallel εἰσὶ δὲ καὶ παράλληλοι Ve paraleldirler deAlso ΕΒ and ΘΓ join them καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ ΘΓ Ayrıca ΕΒ ve ΘΓ onları birleştirirAnd [straights] that join equals and αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ Ve eşit ve paralelleri aynı tarafta bir-

parallels in the same parts τὰ αὐτὰ μέρη ἐπιζευγνύουσαι leştiren doğrularare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσι eşit ve paraleldirler[Also therefore ΕΒ and ΗΘ [καὶ αἱ ΕΒ ΘΓ ἄρα [Yine dolayısıyla ΕΒ ve ΗΘare equal and parallel] ἴσαι τέ εἰσι καὶ παράλληλοι] eşit ve paraleldirler]Therefore a parallelogram is ΕΒΓΘ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ Dolayısıyla ΕΒΓΘ bir paralelkenardırAnd it is equal to ΑΒΓΔ καί ἐστιν ἴσον τῷ ΑΒΓΔ Ve eşittir ΑΒΓΔ paralelkenarınaFor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει Ccediluumlnkuuml onunla aynıΒΓ τὴν ΒΓ ΒΓ tabanı vardırand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve onunla aynıas it it is ΒΓ and ΑΘ ἐστὶν αὐτῷ ταῖς ΒΓ ΑΘ ΒΓ ve ΑΘ paralellerindedirFor the same [reason] then δὶα τὰ αὐτὰ δὴ Aynı sebeple o şimdialso ΕΖΗΘ to it [namely] ΕΒΓΘ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ΕΖΗΘ da ona [yani] ΕΒΓΘ paralelke-is equal ἐστιν ἴσον narınaso that parallelogram ΑΒΓΔ ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον eşittirto ΕΖΗΘ is equal τῷ ΕΖΗΘ ἐστιν ἴσον boumlylece ΑΒΓΔ

ΕΖΗΘ paralelkenarına eşittir

Therefore parallelograms Τὰ ἄρα παραλληλόγραμμα Dolayısıyla paralelkenarlarthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlarda olanlarand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittirler birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Ζ Η

Θ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΒΓ τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ uumlccedilgenlerion the same base ΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ aynı ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΑΔ and ΒΓ ταῖς ΑΔ ΒΓ ΑΔ ve ΒΓ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ΔΒΓ uumlccedilgenine

Suppose has been extended ᾿Εκβεβλήσθω Varsayılsın uzatılmış olduğu

ΑΔ on both sides to Ε and Ζ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε Ζ ΑΔ doğrusunun her iki kenarda Ε veand through Β καὶ διὰ μὲν τοῦ Β Ζ noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΕ ἤχθω ἡ ΒΕ ΓΑ kenarına paraleland through Γ δὶα δὲ τοῦ Γ ΒΕ ccedilizilmiş olsunparallel to ΒΔ τῇ ΒΔ παράλληλος ve Γ noktasındanhas been drawn ΓΖ ἤχθω ἡ ΓΖ ΒΔ kenarına papalel

ΓΖ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΕΒΓΑ and ΔΒΓΖ ἐστὶν ἑκάτερον τῶν ΕΒΓΑ ΔΒΓΖ ΕΒΓΑ ile ΔΒΓΖand they are equal καί εἰσιν ἴσα ve bunlar eşittirfor they are on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι aynıΒΓ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΕΖ ταῖς ΒΓ ΕΖ ΒΓ ve ΕΖ paralellerinde oldukları iccedilinand [it] is καί ἐστι veof the parallelogram ΕΒΓΑ τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ΕΒΓΑ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον mdash ΑΒΓ uumlccedilgenidirfor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει ΑΒ koumlşegeni onu ikiye kestiği iccedilinand of the parallelogram ΔΒΓΖ τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ve ΔΒΓΖ paralelkenarınınhalf ἥμισυ yarısımdashthe triangle ΔΒΓ τὸ ΔΒΓ τρίγωνον mdash ΔΒΓ uumlccedilgenidirfor the diameter ΔΓ cuts it in two ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει ΔΓ koumlşegeni onu ikiye kestiği iccedilin[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση [Ve eşitlerin yarılarıare equal to one another] ἴσα ἀλλήλοις ἐστίν] eşittirler birbirlerine]Therefore equal is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ to the triangle ΔΒΓ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ ΑΒΓ uumlccedilgeni ΔΒΓ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α D

Γ

Ε Ζ

Triangles Τὰ τρίγωνα Uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ίσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuntriangles ΑΒΓ and ΔΕΖ τρίγωνα τὰ ΑΒΓ ΔΕΖ ΑΒΓ ve ΔΕΖ uumlccedilgenlerion equal bases ΒΓ and ΕΖ επὶ ἴσων βάσεων τῶν ΒΓ ΕΖ eşit ΒΓ ve ΕΖ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΖ and ΑΔ ταῖς ΒΖ ΑΔ ΒΖ ve ΑΔ paralellerinde

I say that λέγω ὅτι İddia ediyorum kiequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgeni

CHAPTER ELEMENTS

to triangle ΔΕΖ τῷ ΔΕΖ τριγώνῳ ΔΕΖ uumlccedilgenine

For suppose has been extended ᾿Εκβεβλήσθω γὰρ Ccediluumlnkuuml varsayılsın uzatılmış olduğuΑΔ on both sides to Η and Θ ἡ ΑΔ ἐφ᾿ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η Θ ΑΔ kenarının her iki kenarda Η ve Θand through Β καὶ διὰ μὲν τοῦ Β noktalarınaparallel to ΓΑ τῇ ΓΑ παράλληλος ve Β noktasındanhas been drawn ΒΗ ήχθω ἡ ΒΗ ΓΑ kenarına paraleland through Ζ δὶα δὲ τοῦ Ζ ΒΗ ccedilizilmiş olsunparallel to ΔΕ τῇ ΔΕ παράλληλος ve Ζ noktasındanhas been drawn ΖΘ ήχθω ἡ ΖΘ ΔΕ kenarına paralel

ΖΘ ccedilizilmiş olsun

Therefore a parallelogram παραλληλόγραμμον ἄρα Dolayısıyla birer paralelkenardıris either of ΗΒΓΑ and ΔΕΖΘ ἐστὶν ἑκάτερον τῶν ΗΒΓΑ ΔΕΖΘ ΗΒΓΑ ile ΔΕΖΘand ΗΒΓΑ [is] equal to ΔΕΖΘ καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ ve ΗΒΓΑ eşittir ΔΕΖΘ paralelke-for they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι narınaΒΓ and ΕΖ τῶν ΒΓ ΕΖ eşitand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΓ ve ΕΖ tabanlarındaΒΖ and ΗΘ ταῖς ΒΖ ΗΘ ve aynıand [it] is καί ἐστι ΒΖ ve ΗΘ paralellerinde olduklarıof the parallelogram ΗΒΓΑ τοῦ μὲν ΗΒΓΑ παραλληλογράμμου iccedilinhalf ἥμισυ vemdashthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΗΒΓΑ paralelkenarınınFor the diameter ΑΒ cuts it in two ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει yarısıand of the parallelogram ΔΕΖΘ τοῦ δὲ ΔΕΖΘ παραλληλογράμμου mdashΑΒΓ uumlccedilgenidirhalf ἥμισυ ΑΒ koumlşegeni onu ikiye kestiği iccedilinmdashthe triangle ΖΕΔ τὸ ΖΕΔ τρίγωνον ve ΔΕΖΘ paralelkenarınınfor the diameter ΔΖ cuts it in two ἡ γὰρ ΔΖ δίαμετρος αὐτὸ δίχα τέμνει yarısı[And halves of equals [τὰ δὲ τῶν ἴσων ἡμίση mdash ΖΕΔ uumlccedilgenidirare equal to one another] ἴσα ἀλλήλοις ἐστίν] ΔΖ koumlşegeni onu ikiye kestiği iccedilinTherefore equal is ἴσον ἄρα ἐστὶ [Ve eşitlerin yarılarıthe triangle ΑΒΓ to the triangle ΔΕΖ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ eşittirler birbirlerine]

Dolayısıyla eşittirΑΒΓ uumlccedilgeni ΔΕΖ uumlccedilgenine

Therefore triangles Τὰ ἄρα τρίγωνα Dolayısıyla uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralelerde olanlarare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerinemdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε Ζ

ΘΗ

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΔΒΓ ἴσα τρίγωνα τὰ ΑΒΓ ΔΒΓ ΑΒΓ ve ΔΒΓ eşit uumlccedilgenleribeing on the same base ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı ΒΓ tabanındaand on the same side of ΒΓ καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ ve onun aynı tarafında olan

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose has been joined ΑΔ ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml ΑΔ doğrusunun birleştirilmişolduğu varsayılsın

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΓ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ paraleldir ΑΔ ΒΓ tabanına

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω ccedilizilmiş olduğu varsayılsınthrough the point Α διὰ τοῦ Α σημείου Α noktasındanparallel to the straight ΒΓ τῇ ΒΓ εὐθείᾳ παράλληλος ΒΓ doğrusuna paralelΑΕ ἡ ΑΕ ΑΕ doğrusununand there has been joined ΕΓ καὶ ἐπεζεύχθω ἡ ΕΓ ve birleştirildiği ΕΓ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Eşittir dolayısıylathe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor on the same base ἐπί τε γὰρ τῆς αὐτῆς βάσεώς onunla aynıas it it is ΒΓ ἐστιν αὐτῷ τῆς ΒΓ ΒΓ tabanındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynı paralellerde olduğu iccedilinBut ΑΒΓ is equal to ΔΒΓ ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον Ama ΑΒΓ eşittir ΔΒΓ uumlccedilgenineAlso therefore ΔΒΓ to ΕΒΓ is equal καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ Ve dolayısıyla ΔΒΓ ΕΒΓ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι eşittirwhich is impossible ὅπερ ἐστὶν ἀδύνατον buumlyuumlk kuumlccediluumlğeTherefore is not parallel ΑΕ to ΒΓ οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ ki bu imkansızdırSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι Dolayısıyla paralel değildir ΑΕ ΒΓneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ doğrusunatherefore ΑΔ is parallel to ΒΓ ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος Benzer şekilde o zaman goumlstereceğiz ki

ΑΔ dışındakiler de paralel değildid dolayısıyla ΑΔ ΒΓ doğrusuna paar-

aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on the same base τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα aynı tabandaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve onun aynı tarafında olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

Equal triangles Τὰ ἴσα τρίγωνα Eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler de

Let there be ῎Εστω Verilmiş olsunequal triangles ΑΒΓ and ΓΔΕ ἴσα τρίγωνα τὰ ΑΒΓ ΓΔΕ eşit ΑΒΓ ve ΓΔΕ uumlccedilgenlerion equal bases ΒΓ and ΓΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ ΓΕ eşit ΒΓ ve ΓΕ tabanlarındaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olan

CHAPTER ELEMENTS

I say that λέγω ὅτι İddia ediyorum kithey are also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralellerdedirler de

For suppose ΑΔ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΔ Ccediluumlnkuuml varsayılsın ΑΔ doğrusununbirleştirildiği

I say that λέγω ὅτι İddia ediyorum kiparallel is ΑΔ to ΒΕ παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ paraleldir ΑΔ ΒΕ doğrusuna

For if not Εἰ γὰρ μή Ccediluumlnkuuml eğer değil isesuppose there has been drawn ἤχθω varsayılsın birleştirildiğithrough the point Α διὰ τοῦ Α Α noktasındanparallel to ΒΕ τῇ ΒΕ παράλληλος ΒΕ doğrusuna paralelΑΖ ἡ ΑΖ ΑΖ doğrusununand there has been joined ΖΕ καὶ ἐπεζεύχθω ἡ ΖΕ ve birleştirildiği ΖΕ doğrusununEqual therefore is ἴσον ἄρα ἐστὶ Dolayısıyla eşittirthe triangle ΑΒΓ τὸ ΑΒΓ τρίγωνον ΑΒΓ uumlccedilgenito the triangle ΖΓΕ τῷ ΖΓΕ τριγώνῳ ΖΓΕ uumlccedilgeninefor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι eşitΒΓ and ΓΕ τῶν ΒΓ ΓΕ ΒΓ ve ΓΕ tabanlarındaand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΕ and ΑΖ ταῖς ΒΕ ΑΖ ΒΕ veΑΖ paralellerinde oldukları iccedilinBut the triangle ΑΒΓ ἀλλὰ τὸ ΑΒΓ τρίγωνον Fakat ΑΒΓ uumlccedilgeniis equal to the [triangle] ΔΓΕ ἴσον ἐστὶ τῷ ΔΓΕ [τρίγωνῳ] eşittir ΔΓΕ uumlccedilgeninealso therefore the [triangle] ΔΓΕ καὶ τὸ ΔΓΕ ἄρα [τρίγωνον] ve dolayısıyla ΔΓΕ uumlccedilgeniniis equal to the triangle ΖΓΕ ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ eşittir ΖΓΕ uumlccedilgeninethe greater to the less τὸ μεῖζον τῷ ἐλάσσονι buumlyuumlk kuumlccediluumlğewhich is impossible ὅπερ ἐστὶν ἀδύνατον ki bu imkansızdırTherefore is not parallel ΑΖ to ΒΕ οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ Dolayısıyla paralel değildir ΑΖ ΒΕSimilarly then we shall show that ὁμοίως δὴ δείξομεν ὅτι doğrusunaneither is any other but ΑΔ οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ Benzer şekilde o zaman goumlstereceğiz kitherefore ΑΔ to ΒΕ is parallel ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος ΑΔ dışındakiler de paralel değildir

dolayısıyla ΑΔ ΒΕ doğrusuna par-aleldir

Therefore equal triangles Τὰ ἄρα ἴσα τρίγωνα Dolayısıyla eşit uumlccedilgenlerthat are on equal bases τὰ ἐπὶ ἴσων βάσεων ὄντα eşit tabanlardaand in the same parts καὶ ἐπὶ τὰ αὐτὰ μέρη ve aynı tarafta olanare also in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν aynı paralelerdedirler demdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ Ε

Ζ

If a parallelogram ᾿Εὰν παραλληλόγραμμον Eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgenin

For the parallelogram ΑΒΓΔ Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ Ccediluumlnkuuml ΑΒΓΔ paralelkenarınınas the triangle ΕΒΓ τριγώνῳ τῷ ΕΒΓ ΕΒΓ uumlccedilgeniylemdashsuppose it has the same base ΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ mdashaynı ΒΓ tabanı olduğu varsayılsın

and is in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde oldukları

I say that λέγω ὅτι İddia ediyorum kidouble is διπλάσιόν ἐστι iki katıdırthe parallelogram ΑΒΓΔ τὸ ΑΒΓΔ παραλληλόγραμμον ΑΒΓΔ paralelkenarıof the triangle ΒΕΓ τοῦ ΒΕΓ τριγώνου ΒΕΓ uumlccedilgeninin

For suppose ΑΓ has been joined ᾿Επεζεύχθω γὰρ ἡ ΑΓ Ccediluumlnkuuml varsayılsın ΑΓ doğrusununbirleştirildiği

Equal is the triangle ΑΒΓ ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον Eşittir ΑΒΓ uumlccedilgenito the triangle ΕΒΓ τῷ ᾿ΕΒΓ τριγώνῳ ΕΒΓ uumlccedilgeninefor it is on the same base as it ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ onunla aynıΒΓ τῆς ΒΓ ΒΓ tabanına sahipand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ve aynıΒΓ and ΑΕ ταῖς ΒΓ ΑΕ ΒΓ ve ΑΕ paralelerinde olduğu iccedilinBut the parallelogram ΑΒΓΔ ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον Fakat ΑΒΓΔ paralelkenarıis double of the triangle ΑΒΓ διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου iki katıdır ΑΒΓ uumlccedilgenininfor the diameter ΑΓ cuts it in two ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει ΑΓ koumlşegeni onu ikiye kestiğindenso that the parallelogram ΑΒΓΔ ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον boumlylece ΑΒΓΔ paralelkenarı daalso of the triangle ΕΒΓ is double ᾿καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον ΕΒΓ uumlccedilgeninin iki katıdır

Therefore if a parallelogram ᾿Εὰν ἄρα παραλληλόγραμμον Dolayısıyla eğer bir paralelkenarhave the same base as a triangle τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν bir uumlccedilgenle aynı tabana sahipseand be in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ ve aynı paralelerdeysedouble is διπλάσιόν ἐστί iki katıdırthe parallelogram of the triangle τὸ παραλληλόγραμμον τοῦ τριγώνου paralelkenar uumlccedilgeninmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε

To the given triangle equal Τῷ δοθέντι τριγώνῳ ἴσον Verilen bir uumlccedilgene eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarıin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlzkenar accedilıda inşa etmek

Let be ῎Εστω Verilenthe given triangle ΑΒΓ τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ uumlccedilgen ΑΒΓand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ ve verilen duumlzkenar accedilı Δ olsun

It is necessary then δεῖ δὴ Şimdi gerklidirto the triangle ΑΒΓ equal τῷ ΑΒΓ τριγώνῳ ἴσον ΑΒΓ uumlccedilgenine eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenarınin the rectilineal angle Δ ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ Δ duumlzkenar accedilısına inşa edilmesi

Suppose ΒΓ has been cut in two at Ε Τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε Varsayılsın ΒΓ kenarının Ε nok-and there has been joined ΑΕ καὶ ἐπεζεύχθω ἡ ΑΕ tasında ikiye kesildiğiand there has been constructed καὶ συνεστάτω ve ΑΕ doğrusunun birleştirildiğion the straight ΕΓ πρὸς τῇ ΕΓ εὐθείᾳ ve inşa edildiğiand at the point Ε on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε ΕΓ doğrusundato angle Δ equal τῇ Δ γωνίᾳ ἴση ve uumlzerindekiΕ noktasındaΓΕΖ ἡ ὑπὸ ΓΕΖ Δ accedilısına eşitalso through Α parallel to ΕΓ καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ΓΕΖ accedilısının

CHAPTER ELEMENTS

suppose ΑΗ has been drawn ἤχθω ἡ ΑΗ ayrıca Α noktasından ΕΓ doğrusunaand through Γ parallel to ΕΖ διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος paralelsuppose ΓΗ has been drawn ἤχθω ἡ ΓΗ ΑΗ doğrusunun ccedilizilmiş olduğutherefore a parallelogram is ΖΕΓΗ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ varsayılsın

ve Γ noktasından ΕΖ doğrusuna par-alel

ΓΗ doğrusunun ccedilizilmiş olduğuvarsayılsın

dolayısıyla ΖΕΓΗ bir paralelkenardır

And since equal is ΒΕ to ΕΓ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ Ve eşit olduğundan ΒΕ ΕΓequal is also ἴσον ἐστὶ καὶ doğrusunatriangle ΑΒΕ to triangle ΑΕΓ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ eşittirfor they are on equal bases ἐπί τε γὰρ ἴσων βάσεών εἰσι ΑΒΕ uumlccedilgeni de ΑΕΓ uumlccedilgenineΒΕ and ΕΓ τῶν ΒΕ ΕΓ tabanlarıand in the same parallels καὶ ἐν ταῖς αὐταῖς παραλλήλοις ΒΕ ve ΕΓ eşitΒΓ and ΑΗ ταῖς ΒΓ ΑΗ ve aynıdouble therefore is διπλάσιον ἄρα ἐστὶ ΒΓ ve ΑΗ paralelerinde oldukları iccedilintriangle ΑΒΓ of triangle ΑΕΓ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου iki katıdır dolayısıylaalso is ἔστι δὲ καὶ ΑΒΓ uumlccedilgeni ΑΕΓ uumlccedilgenininparallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον ayrıcadouble of triangle ΑΕΓ διπλάσιον τοῦ ΑΕΓ τριγώνου ΖΕΓΗ paralelkenarıfor it has the same base as it βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει iki katıdır ΑΕΓ uumlccedilgenininand καὶ onunla aynı tabanı olduğuis in the same parallels as it ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις vetherefore is equal ἴσον ἄρα ἐστὶ onunla aynı paralellerde olduğu iccedilinthe parallelogram ΖΕΓΗ τὸ ΖΕΓΗ παραλληλόγραμμον dolayısıyla eşittirto the triangle ΑΒΓ τῷ ΑΒΓ τριγώνῳ ΖΕΓΗ paralelkenarıAnd it has angle ΓΕΖ καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ΑΒΓ uumlccedilgenineequal to the given Δ ἴσην τῇ δοθείσῃ τῇ Δ Ve onun ΓΕΖ accedilısı

eşittir verilen Δ accedilısına

Therefore to the given triangle ΑΒΓ Τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ Dolayısıyla verilen ΑΒΓ uumlccedilgenineequal ἴσον eşita parallelogram has been constructed παραλληλόγραμμον συνέσταται bir paralelkenarΖΕΓΗ τὸ ΖΕΓΗ ΖΕΓΗ inşa edilmiş olduin the angle ΓΕΖ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ ΓΕΖ aşısındawhich is equal to Δ ἥτις ἐστὶν ἴση τῇ Δ Δ aşısına eşit olanmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

ΖΑ

Β

Δ

ΓΕ

Η

Of any parallelogram Παντὸς παραλληλογράμμου Herhangi bir paralelkenarınof the parallelograms about the diam- τῶν περὶ τὴν διάμετρον παραλληλο- koumlşegeni etrafındaki

eter γράμμων paralelkenarlarınthe complements τὰ παραπληρώματα tuumlmleyenleriare equal to one another ἴσα ἀλλήλοις ἐστίν eşittir birbirlerine

Let there be ῎Εστω Verilmiş olsuna parallelogram ΑΒΓΔ παραλληλόγραμμον τὸ ΑΒΓΔ bir ΑΒΓΔ paralelkenarıand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve onun ΑΓ koumlşegeniand about ΑΓ περὶ δὲ τὴν ΑΓ ve ΑΓ etrafındalet be parallelograms παραλληλόγραμμα μὲν ἔστω paralelkenarlarΕΘ and ΖΗ τὰ ΕΘ ΖΗ ΕΘ ve ΖΗ

and the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenleriΒΚ and ΚΔ τὰ ΒΚ ΚΔ ΒΚ ile ΚΔ

I say that λέγω ὅτι İddia ediyorumequal is the complement ΒΚ ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα eşittir ΒΚ tuumlmleyenito the complement ΚΔ τῷ ΚΔ παραπληρώματι ΚΔ tuumlmleyenine

For since a parallelogram is ᾿Επεὶ γὰρ παραλληλόγραμμόν ἐστι Ccediluumlnkuuml bir paralelkenar olduğundanΑΒΓΔ τὸ ΑΒΓΔ ΑΒΓΔand its diameter ΑΓ διάμετρος δὲ αὐτοῦ ἡ ΑΓ ve ΑΓ onun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΒΓ to triangle ΑΓΔ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ ΑΒΓ uumlccedilgeni ΑΓΔ uumlccedilgenineMoreover since a parallelogram is πάλιν ἐπεὶ παραλληλόγραμμόν ἐστι Dahası bir paralelkenar olduğundanΕΘ τὸ ΕΘ ΕΘand its diameter ΑΚ διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ ΑΚonun koumlşegeniequal is ἴσον ἐστὶ eşittirtriangle ΑΕΚ to triangle ΑΘΚ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ΑΕΚ uumlccedilgeni ΑΘΚuumlccedilgenineThen for the same [reasons] also διὰ τὰ αὐτὰ δὴ καὶ Şimdi aynı nedenletriangle ΚΖΓ to ΚΗΓ is equal τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον ΚΖΓ eşittir ΚΗΓ uumlccedilgenineSince then triangle ΑΕΚ ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον O zaman ΑΕΚis equal to triangle ΑΘΚ τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον eşit olduğundan ΑΘΚ uumlccedilgenineand ΚΖΓ to ΚΗΓ τὸ δὲ ΚΖΓ τῷ ΚΗΓ ve ΚΖΓ ΚΗΓ uumlccedilgeninetriangle ΑΕΚ with ΚΗΓ τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ΑΕΚ ile ΚΗΓ uumlccedilgenleriis equal ἴσον ἐστὶ eşittirlto triangle ΑΘΚ with ΚΖΓ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ ΑΘΚ ile ΚΖΓ uumlccedilgenlerinealso is triangle ΑΒΓ as a whole ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ayrıca ΑΒΓ uumlccedilgeninin tuumlmuumlequal to ΑΔΓ as a whole ὅλῳ τῷ ΑΔΓ ἴσον eşittir ΑΔΓ uumlccedilgeninin tuumlmuumlnetherefore the complement ΒΚ remain- λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα dolayısıyla geriye kalan ΒΚ tuumlmleyeni

ing λοιπῷ τῷ ΚΔ παραπληρώματί geriye kalan ΚΔ tuumlmleyenineto the complement ΚΔ remaining ἐστιν ἴσον eşittiris equal

Therefore of any parallelogram area Παντὸς ἄρα παραλληλογράμμου χωρίου Dolayısıyla herhangi bir paralelke-of the about-the-diameter τῶν περὶ τὴν διάμετρον narınparallelograms παραλληλογράμμων koumlşegeni etrafındakithe complements τὰ παραπληρώματα paralelkenarlarınare equal to one another ἴσα ἀλλήλοις ἐστίν tuumlmleyenlerimdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι eşittir birbirlerine

mdashgoumlsterilmesi gereken tam buydu

Β

Α Δ

Γ

Ε Z

Θ

Η

Κ

Along the given straight Παρὰ τὴν δοθεῖσαν εὐθεῖαν Verilen bir doğru boyuncaequal to the given triangle τῷ δοθέντι τριγώνῳ ἴσον verilen bir uumlccedilgene eşitto apply a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralel kenarı yerleştirmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen bir duumlz kenar accedilıda

Let be ῎Εστω Verilen doğru ΑΒthe given straight ΑΒ ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ ve verilen uumlccedilgen Γ

Here Euclid can use two letters without qualification for a par-allelogram because they are not unqualified in the Greek they takethe neuter article while a line takes the feminine article

This is Heathrsquos translation The Greek does not require any-

thing corresponding to lsquoso-rsquo The LSJ lexicon [] gives the presentproposition as the original geometrical use of παραπλήρωμαmdashothermeanings are lsquoexpletiversquo and a certain flowering herb

CHAPTER ELEMENTS

and the given triangle Γ τὸ δὲ δοθὲν τρίγωνον τὸ Γ ve verlen duumlzkenar accedilı Δ olsunand the given rectilineal angle Δ ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ

It is necessary then δεῖ δὴ Şimdi gereklidiralong the given straight ΑΒ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ verilen ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον Γ uumlccedilgenine eşitto appy a parallelogram παραλληλόγραμμον παραβαλεῖν bir paralelkenarıin an equal to the angle Δ ἐν ἴσῃ τῇ Δ γωνίᾳ Δ accedilısında yerleştirmek

Suppose has been constructed Συνεστάτω Varsayılsın inşa edildiğiequal to triangle Γ τῷ Γ τριγώνῳ ἴσον Γ uumlccedilgenine eşita parallelogram ΒΕΖΗ παραλληλόγραμμον τὸ ΒΕΖΗ bir ΒΕΖΗ paralelkenarınınin angle ΕΒΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ ΕΒΗ accedilısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olanΔ accedilısınaand let it be laid down καὶ κείσθω ve oumlyle yerleştirilmiş olsun kiso that on a straight is ΒΕ ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ bir doğruda kalsın ΒΕwith ΑΒ τῇ ΑΒ ΑΒ ileand suppose has been drawn through καὶ διήχθω ve ccedilizilmiş olsunΖΗ to Θ ἡ ΖΗ ἐπὶ τὸ Θ ΖΗ dogrusundan Θ noktasınaand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΗ and ΕΖ ὁποτέρᾳ τῶν ΒΗ ΕΖ paralel olan ΒΗ ve ΕΖ doğrularındansuppose there has been drawn παράλληλος ἤχθω ἡ ΑΘ birineΑΘ καὶ ἐπεζεύχθω ἡ ΘΒ ccedilizilmiş olsunand suppose there has been joined ΑΘΘΒ ve birleştirilmiş olsun

ΘΒ

And since on the parallels ΑΘ and ΕΖ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ ΕΖ Ve ΑΘ ile ΕΖ paralellerinin uumlzerinefell the straight ΘΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ duumlştuumlğuumlnden ΘΖ doğrusuthe angles ΑΘΖ and ΘΖΕ αἱ ἄρα ὑπὸ ΑΘΖ ΘΖΕ γωνίαι ΑΘΖ ve ΘΖΕ accedilılarıare equal to two rights δυσὶν ὀρθαῖς εἰσιν ἴσαι eşittir iki dik accedilıyaTherefore ΒΘΗ and ΗΖΕ αἱ ἄρα ὑπὸ ΒΘΗ ΗΖΕ Dolayısıyla ΒΘΗ veΗΖΕare less than two rights δύο ὀρθῶν ἐλάσσονές εἰσιν kuumlccediluumlktuumlr iki dik accedilıdanAnd [straights] from [angles] that αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς Ve kuumlccediluumlk olanlardan

are less ἄπειρον ἐκβαλλόμεναι iki dik accedilıdanthan two rights συμπίπτουσιν uzatıldıklarında sonsuzaextended to the infinite αἱ ΘΒ ΖΕ ἄρα ἐκβαλλόμεναι birbirlerine duumlşerler doğrularfall together συμπεσοῦνται Dolayısıyla ΘΒ ve ΖΕ uzatılırsaTherefore ΘΒ and ΖΕ extended birbirlerine duumlşerlerfall together

Suppose they have been extended ἐκβεβλήσθωσαν Varsayılsın uzatıldıklarıand they have fallen together at Κ καὶ συμπιπτέτωσαν κατὰ τὸ Κ ve Κ noktasında kesiştikleriand through the point Κ καὶ διὰ τοῦ Κ σημείου ve Κ noktasındanparallel to either of ΕΑ and ΖΘ ὁποτέρᾳ τῶν ΕΑ ΖΘ παράλληλος paralel olan ΕΑ veya ΖΘ doğrusunasuppose has been drawn ΚΛ ἤχθω ἡ ΚΛ ccedilizilmiş olsun ΚΛand suppose have been extended ΘΑ καὶ ἐκβεβλήσθωσαν αἱ ΘΑ ΗΒ ve uzatılmış olsunlarΘΑ ve ΗΒ doğru-

and ΗΒ ἐπὶ τὰ Λ Μ σημεῖα larıto the points Λ and Μ Λ ve Μ noktalarından

A parallelogram therefore is ΘΛΚΖ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ Bir paralelkenardır dolayısıyla ΘΛΚΖa diameter of it is ΘΚ διάμετρος δὲ αὐτοῦ ἡ ΘΚ ve onun koumlşegeni ΘΚand about ΘΚ [are] περὶ δὲ τὴν ΘΚ ve ΘΚ etrafındadırthe parallelograms ΑΗ and ΜΕ παραλληλόγραμμα μὲν τὰ ΑΗ ΜΕ ΑΗ ve ΜΕ paralelkenarlarıand the so-called complements τὰ δὲ λεγόμενα παραπληρώματα ve bunların tuumlmleyenlerisΛΒ and ΒΖ τὰ ΛΒ ΒΖ ΛΒ ileΒΖequal therefore is ΛΒ to ΒΖ ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ eşittirler dolayısıyla ΛΒ ile ΒΖBut ΒΖ to triangle Γ is equal ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον tuumlmleyenlerineAlso therefore ΛΒ to Γ is equal καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον Ama ΒΖ Γ uumlccedilgenine eşittirAnd since equal is καὶ ἐπεὶ ἴση ἐστὶν Dolaysısıyla ΛΒ da Γ uumlccedilgenine eşittirangle ΗΒΕ to ΑΒΜ ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ Ve eşit olduğundanbut ΗΒΕ to Δ is equal ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση ΗΒΕ ΑΒΜ accedilısınaalso therefore ΑΒΜ to Δ καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ fakat ΗΒΕ Δ accedilısına eşit

is equal ἐστὶν ἴση dolayısıyla ΑΒΜ de Δ accedilısınaeşittir

Therefore along the given straight Παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν Dolaysısyla verilen birΑΒ τὴν ΑΒ ΑΒ doğrusu boyuncaequal to the given triangle Γ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον verilen bir Γ uumlccedilgenine eşita parallelogram has been applied παραλληλόγραμμον παραβέβληται birΛΒ τὸ ΛΒ ΛΒ paralelkenarı yerleştirilmiş olduin the angle ΑΒΜ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ ΑΒΜ aşısındawhich is equal to Δ ἥ ἐστιν ἴση τῇ Δ eşit olan Δ accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

ΕΖ

Θ

Η

Κ

Λ

Μ

ΔΓ

To the given rectilineal [figure] equal Τῷ δοθέντι εὐθυγράμμῳ ἴσον Verilen bir duumlzkenar [figuumlre] eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given rectilineal angle ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ verilen duumlzkenar accedilıda

Let be ῎Εστω Verilmiş olsunthe given rectilineal [figure] ΑΒΓΔ τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ ΑΒΓΔ duumlzkenar [figuumlruuml]and the given rectilineal angle Ε ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε ve duumlzkenar Ε accedilısı

It is necessary then δεῖ δὴ Gereklidir şimdito the rectilineal ΑΒΓΔ equal τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον ΑΒΓΔ duumlzkenarına eşita parallelogram to construct παραλληλόγραμμον συστήσασθαι bir paralelkenar inşa etmekin the given angle Ε ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε verilen Ε accedilısında

Suppose has been joined ΔΒ ᾿Επεζεύχθω ἡ ΔΒ Birleştirilmiş olduğu ΔΒ doğrusununand suppose has been constructed καὶ συνεστάτω ve inşa edilmiş olsunequal to the triangle ΑΒΔ τῷ ΑΒΔ τριγώνῳ ἴσον ΑΒΔ uumlccedilgenine eşita parallelogram ΖΘ παραλληλόγραμμον τὸ ΖΘ bir ΖΘ paralelkenarıin the angle ΘΚΖ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ ΘΚΖ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısınaand suppose there has been applied καὶ παραβεβλήσθω ve yerleştirilmiş olsunalong the straight ΗΘ παρὰ τὴν ΗΘ εὐθεῖαν ΗΘ doğrusu boyunca equal to triangle ΔΒΓ τῷ ΔΒΓ τριγώνῳ ἴσον ΔΒΓ uumlccedilgenine eşita parallelogram ΗΜ παραλληλόγραμμον τὸ ΗΜ bir ΗΜ paralelkenarıin the angle ΗΘΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ ΗΘΜ accedilısındawhich is equal to Ε ἥ ἐστιν ἴση τῇ Ε eşit olan Ε accedilısına

And since angle Ε καὶ ἐπεὶ ἡ Ε γωνία Ve Ε accedilısıto either of ΘΚΖ and ΗΘΜ ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ ΗΘΜ ΘΚΖ ve ΗΘΜ accedilılarının her birineis equal ἐστιν ἴση eşit olduğundantherefore also ΘΚΖ to ΗΘΜ καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ΘΚΖ da ΗΘΜ accedilısınais equal ἐστιν ἴση eşittirLet ΚΘΗ be added in common κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ Eklenmiş olsun ΚΘΗ ortak olaraktherefore ΖΚΘ and ΚΘΗ αἱ ἄρα ὑπὸ ΖΚΘ ΚΘΗ dolayısıyla ΖΚΘ ve ΚΘΗto ΚΘΗ and ΗΘΜ ταῖς ὑπὸ ΚΘΗ ΗΘΜ ΚΘΗ ve ΗΘΜ accedilılarınaare equal ἴσαι εἰσίν eşittirlerBut ΖΚΘ and ΚΘΗ ἀλλ᾿ αἱ ὑπὸ ΖΚΘ ΚΘΗ Fakat ΖΚΘ ve ΚΘΗare equal to two rights δυσὶν ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıyatherefore also ΚΘΗ and ΗΘΜ καὶ αἱ ὑπὸ ΚΘΗ ΗΘΜ ἄρα dolayısıyla ΚΘΗ ve ΗΘΜ accedilılarıdaare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittirler iki dik accedilıya

CHAPTER ELEMENTS

Then to some straight ΗΘ πρὸς δή τινι εὐθεῖᾳ τῇ ΗΘ Şimdi bir ΗΘ doğrusunaand at the same point Θ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ ve aynı Θ noktasındatwo straights ΚΘ and ΘΜ δύο εὐθεῖαι αἱ ΚΘ ΘΜ iki ΚΘ ve ΘΜ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας komşu accedilılarımake equal to two rights δύο ὀρθαῖς ἴσας ποιοῦσιν iki dik accedilıya eşit yaparIn a straight then are ΚΘ and ΘΜ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ O zaman bir doğrudadır ΚΘ ve ΘΜand since on the parallels ΚΜ and ΖΗ καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ ΖΗ ve ΚΜ ve ΖΗ paralelleri uumlzerinefell the straight ΘΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ duumlştuumlğuumlnden ΘΗ doğrusuthe alternate angles ΜΘΗ and ΘΗΖ αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ ΘΗΖ ters ΜΘΗ ve ΘΗΖ accedilılarıare equal to one another ἴσαι eşittir birbirineLet ΘΗΛ be added in common ἀλλήλαις εἰσίν eklenmiş olsun ΘΗΛ ortak olaraktherefore ΜΘΗ and ΘΗΛ κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ dolayısıyla ΜΘΗ ve ΘΗΛto ΘΗΖ and ΘΗΛ αἱ ἄρα ὑπὸ ΜΘΗ ΘΗΛ ταῖς ὑπὸ ΘΗΖ ΘΗΖ ve ΘΗΛ accedilılarınaare equal ΘΗΛ eşittirlerBut ΜΘΗ and ΘΗΛ ίσαι εἰσιν Fakat ΜΘΗ ve ΘΗΛare equal to two rights αλλ᾿ αἱ ὑπὸ ΜΘΗ ΘΗΛ eşittirler iki dik accedilıyatherefore also ΘΗΖ and ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν dolayısıyla ΘΗΖ ve ΘΗΛ daare equal to two rights καὶ αἱ ὑπὸ ΘΗΖ ΘΗΛ ἄρα eşittirler iki dik accedilıyatherefore on a straight are ΖΗ and δύο ὀρθαῖς ἴσαι εἰσίν dolaysısyla bir doğru uumlzerindedir ΖΗ

ΗΛ ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ ve ΗΛAnd since ΖΚ to ΘΗ καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ Ve olduğundan ΖΚ ΘΗ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paralelbut also ΘΗ to ΜΛ ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ ve de ΘΗ ΜΛ doğrusunatherefore also ΚΖ to ΜΛ καὶ ἡ ΚΖ ἄρα τῇ ΜΛ dolayısıyla ΚΖ da ΜΛ doğrusunais equal and parallel ἴση τε καὶ παράλληλός ἐστιν eşit ve paraleldirand join them καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ve birleştirir onları ΚΜ ile ΖΛ ki bun-ΚΜ and ΖΛ which are straights ΚΜ ΖΛ larda doğrulardırtherefore also ΚΜ and ΖΛ καὶ αἱ ΚΜ ΖΛ ἄρα dolayısıyla ΚΜ ve ΖΛ daare equal and parallel ἴσαι τε καὶ παράλληλοί εἰσιν eşit ve paraleldirlera parallelogram therefore is ΚΖΛΜ παραλληλόγραμμον ἄρα ἐστὶ τὸ dolayısıyla ΚΖΛΜ bir paralelkenardırAnd since equal is ΚΖΛΜ Ve eşit olduğundantriangle ΑΒΔ καὶ ἐπεὶ ἴσον ἐστὶ ΑΒΔ uumlccedilgenito the parallelogram ΖΘ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλ- ΖΘ paralelkenarınaand ΔΒΓ to ΗΜ ληλογράμμῳ ve ΔΒΓ ΗΜ paralelkenarınatherefore as a whole τὸ δὲ ΔΒΓ τῷ ΗΜ dolayısısyla bir buumltuumln olarakthe rectilineal ΑΒΓΔ ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ΑΒΓΔ duumlzkenarıto parallelogram ΚΖΛΜ as a whole ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ bir buumltuumln olarak ΚΖΛΜ paralelke-is equal ἐστὶν ἴσον narına

eşittir

Therefore to the given rectilineal [fig- Τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ Dolayısıyla verilen duumlzkenar ΑΒΓΔure] ΑΒΓΔ equal ἴσον figuumlruumlne eşit

a parallelogram has been constructed παραλληλόγραμμον συνέσταται bir ΚΖΛΜ paralelkenarı inşa edilmişΚΖΛΜ τὸ ΚΖΛΜ olduin the angle ΖΚΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ ΖΚΜ accedilısındawhich is equal to the given Ε ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε eşit olan verilmiş Ε accedilısınamdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

Β

Α

Ε

Ζ

Θ

Η

Κ

Λ

Μ

Δ

Γ

On the given straight Απὸ τῆς δοθείσης εὐθείας Verilen bir doğrudato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Let be ῎Εστω Verilmiş olsunthe given straight ΑΒ ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ ΑΒ doğrusu

It is required then δεῖ δὴ Şimdi gereklidiron the straight ΑΒ ἀπὸ τῆς ΑΒ εὐθείας ΑΒ doğrusundato set up a square τετράγωνον ἀναγράψαι bir kare kurmak

Suppose there has been drawn ῎Ηχθω Ccedilizilmiş olsunto the straight ΑΒ τῇ ΑΒ εὐθείᾳ ΑΒ doğrusundaat the point Α of it ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α onun Α noktasındaat a right πρὸς ὀρθὰς dik accedilıdaΑΓ ἡ ΑΓ ΑΓand suppose there has been laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΑΒ τῇ ΑΒ ἴση ΑΒ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand through the point Δ καὶ διὰ μὲν τοῦ Δ σημείου ve Δ noktasındanparallel to ΑΒ τῇ ΑΒ παράλληλος ΑΒ doğrusuna paralelsuppose there has been drawn ΔΕ ἤχθω ἡ ΔΕ ccedilizilmiş olsun ΔΕand through the point Β διὰ δὲ τοῦ Β σημείου ve Β noktasındanparallel to ΑΔ τῇ ΑΔ παράλληλος ΑΔ doğrusuna paralelsuppose there has been drawn ΒΕ ἤχθω ἡ ΒΕ ΒΕ ccedilizilmiş olsun

A parallelogram therefore is ΑΔΕΒ παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒequal therefore is ΑΒ to ΔΕ ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ Bir paralelkenardır dolayısıylaand ΑΔ to ΒΕ ἡ δὲ ΑΔ τῇ ΒΕ ΑΔΕΒBut ΑΒ to ΑΔ is equal ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση eşittir dolayısıyla ΑΒ ΔΕ doğrusunaTherefore the four αἱ τέσσαρες ἄρα ve ΑΔ ΒΕ doğrusunaΒΑ ΑΔ ΔΕ and ΕΒ αἱ ΒΑ ΑΔ ΔΕ ΕΒ Ama ΑΒ ΑΔ doğrusuna eşittirare equal to one another ἴσαι ἀλλήλαις εἰσίν Dolaysısyla şu doumlrduumlequilateral therefore ἰσόπλευρον ἄρα ΒΑ ΑΔ ΔΕ ve ΕΒis the parallelogram ΑΔΕΒ ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον birbirlerine eşittirler

eşkenardır dolayısıylaΑΔΕΒ paralelkenarı

I say then that λέγω δή ὅτι Şimdi iddia ediyorum kiit is also right-angled καὶ ὀρθογώνιον aynı zamanda dik accedilılıdır

For since on the parallels ΑΒ and ΔΕ ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ ΔΕ Ccediluumlnkuuml ΑΒ ve ΔΕ paralellerinin uumlzer-fell the straight ΑΔ εὐθεῖα ἐνέπεσεν ἡ ΑΔ inetherefore the angles ΒΑΔ and ΑΔΕ αἱ ἄρα ὑπὸ ΒΑΔ ΑΔΕ γωνίαι duumlştuumlğuumlnden ΑΔ doğrusuare equal to two rights δύο ὀρθαῖς ἴσαι εἰσίν eşittir dolaysıyla ΒΑΔ ve ΑΔΕAnd ΒΑΔ is right ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ iki dik accedilıyaright therefore is ΑΔΕ ορθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ Ve ΒΑΔ diktirAnd of parallelogram areas τῶν δὲ παραλληλογράμμων χωρίων diktir dolayısıyla ΑΔΕthe opposite sides and angles αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι Ve paralelkenar alanlarınare equal to one another ἴσαι ἀλλήλαις εἰσίν karşıt kenar ve accedilılarıRight therefore is either ὀρθὴ ἄρα καὶ ἑκατέρα eşittir birbirlerineof the opposite angles ΑΒΕ and ΒΕΔ τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ ΒΕΔ Diktir dolayısıyla her birright-angled therefore is ΑΔΕΒ γωνιῶν karşıt accedilı ΑΒΕ ve ΒΕΔAnd it was shown also equilateral ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ dik accedilılıdır dolayısıyla ΑΔΕΒ

ἐδείχθη δὲ καὶ ἰσόπλευρον Ve goumlsterilmişti ki eşkenardır da

A square therefore it is Τετράγωνον ἄρα ἐστίν Bir karedir dolayısıyla oand it is on the straight ΑΒ καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ve o ΑΒ doğrusu uumlzerineset up ἀναγεγραμμένον kurulmuşturmdashjust what it was necessary to do ὅπερ ἔδει ποιῆσαι mdashyapılması gereken tam buydu

CHAPTER ELEMENTS

ΒΑ

ΕΔ

Γ

In right-angled triangles ᾿Εν τοῖς ὀρθογωνίοις τριγώνοις Dik accedilılı uumlccedilgenlerdethe square on the side that subtends τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

the right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides that con- τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

tain the right angle χουσῶν πλευρῶν τετραγώνοις ilere

Let be ῎Εστω Verilmiş olsuna right-angled triangle ΑΒΓ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ dik accedilılı bir ΑΒΓ uumlccedilgenihaving the angle ΒΑΓ right ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν ΒΑΓ accedilısı dik olan

I say that λέγω ὅτι İddia ediyorum kithe square on ΓΒ τὸ ἀπὸ τῆς ΒΓ τετράγωνον ΓΒ uumlzerindeki kareis equal ἴσον ἐστὶ eşittirto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις ΒΑ ve ΑΓ uumlzerlerindeki karelere

For suppose there has been set up Αναγεγράφθω γὰρ Ccediluumlnkuuml kurulmuş olsunon ΒΓ ἀπὸ μὲν τῆς ΒΓ ΒΓ uumlzerindea square ΒΔΕΓ τετράγωνον τὸ ΒΔΕΓ bir ΒΔΕΓ karesiand on ΒΑ and ΑΓ ἀπὸ δὲ τῶν ΒΑ ΑΓ ve ΒΑ ile ΑΓ uumlzerlerindeΗΒ and ΘΓ τὰ ΗΒ ΘΓ ΗΒ ve ΘΓand through Α καὶ διὰ τοῦ Α ve Α noktasındanparallel to either of ΒΔ and ΓΕ ὁποτέρᾳ τῶν ΒΔ ΓΕ παράλληλος ΒΔ ve ΓΕ doğrularına paralel olansuppose ΑΛ has been drawn ἤχθω ἡ ΑΛ1 ΑΛ ccedilizilmiş olsunand suppose have been joined καὶ ἐπεζεύχθωσαν ve birleştirilmiş olsunΑΔ and ΖΓ αἱ ΑΔ ΖΓ ΑΔ ve ΖΓ

And since right is καὶ ἐπεὶ ὀρθή ἐστιν Ve dik olduğundaneither of the angles ΒΑΓ and ΒΑΗ ἑκατέρα τῶν ὑπὸ ΒΑΓ ΒΑΗ γωνιῶν ΒΑΓ ve ΒΑΗ accedilılarının her birion some straight ΒΑ πρὸς δή τινι εὐθείᾳ τῇ ΒΑ bir ΒΑ doğrusundato the point Α on it καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α uumlzerindeki Α noktasınatwo straights ΑΓ and ΑΗ δύο εὐθεῖαι αἱ ΑΓ ΑΗ ΑΓ ve ΑΗ doğrularınot lying in the same parts μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι aynı tarafta kalmayanthe adjacent angles τὰς ἐφεξῆς γωνίας bitişik accedilılarmake equal to two rights δυσὶν ὀρθαῖς ἴσας ποιοῦσιν oluştururlar eşit iki dik accedilıyaon a straight therefore is ΓΑ with ἐπ᾿ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ bir doğrudadır dolayısısyla ΓΑ ile ΑΗ

ΑΗ διὰ τὰ αὐτὰ δὴ Sonra aynı nedenleThen for the same [reason] καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾿ εὐθείας ΒΑ ile ΑΘ da bir doğrudadıralso ΒΑ with ΑΘ is on a straight καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ ΔΒΓ ΖΒΑ accedilısınaangle ΔΒΓ to angle ΖΒΑ ὀρθὴ γὰρ ἑκατέρα her ikiside diktirfor either is right κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ eklenmiş olsun ΑΒΓ her ikisine delet ΑΒΓ be added in common ὅλη ἄρα ἡ ὑπὸ ΔΒΑ dolayısıyla ΔΒΑ accedilısının tamamıtherefore ΔΒΑ as a whole ὅλῃ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısının tamamınato ΖΒΓ as a whole ἐστιν ἴση eşittiris equal καὶ ἐπεὶ ἴση ἐστὶν Ve eşit olduğundanAnd since equal is ἡ μὲν ΔΒ τῇ ΒΓ ΔΒ ΒΓ doğrusuna

Heibergrsquos text [ p ] has Δ for Λ at this place and else-where (though not in the diagram) Probably this is a compositorrsquos

mistake owing to the similarity in appearance of the two lettersespecially in the font used

ΔΒ to ΒΓ ἡ δὲ ΖΒ τῇ ΒΑ ve ΖΒ ΒΑ doğrusunaand ΖΒ to ΒΑ δύο δὴ αἱ ΔΒ ΒΑ ΔΒ ve ΒΑ ikilisithe two ΔΒ and ΒΑ δύο ταῖς ΖΒ ΒΓ ΖΒ ve ΒΓ ikilisine

to the two ΖΒ and ΒΓ ἴσαι εἰσὶν eşittirlerare equal ἑκατέρα ἑκατέρᾳ her biri birineeither to either καὶ γωνία ἡ ὑπὸ ΔΒΑ ve ΔΒΑ accedilısıand angle ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ΖΒΓ accedilısınato angle ΖΒΓ ἴση eşittiris equal βάσις ἄρα ἡ ΑΔ dolayısıyla ΑΛ tabanıtherefore the base ΑΛ βάσει τῇ ΖΓ ΖΓ tabanınato the base ΖΓ [ἐστιν] ἴση eşittir[is] equal καὶ τὸ ΑΒΔ τρίγωνον ve ΑΒΔ uumlccedilgeniand the triangle ΑΒΔ τῷ ΖΒΓ τριγώνῳ ΖΒΓ uumlccedilgenineto the triangle ΖΒΓ ἐστὶν ἴσον eşittiris equal καί [ἐστι] τοῦ μὲν ΑΒΔ τριγώνου ve ΑΒΔ uumlccedilgenininand of the triangle ΑΒΔ διπλάσιον τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı iki katıdırthe parallelogram ΒΛ is double βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ aynı ΒΛ tabanları olduğufor they have the same base ΒΛ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΒΔ ΑΛ ΒΔ ve ΑΛ parallerinde oldukları iccedilinΒΔ and ΑΛ τοῦ δὲ ΖΒΓ τριγώνου ve ΖΒΓ uumlccedilgenininand of the triangle ΖΒΓ διπλάσιον τὸ ΗΒ τετράγωνον ΗΒ karesi iki katıdırthe square ΗΒ is double βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι yine aynıfor again they have the same base τὴν ΖΒ ΖΒ tabanları olduğuΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ve aynıand are in the same parallels ταῖς ΖΒ ΗΓ ΖΒ ve ΗΓ parallerinde oldukları iccedilinΖΒ and ΗΓ [τὰ δὲ τῶν ἴσων [Ve eşitlerin[And of equals διπλάσια ἴσα ἀλλήλοις ἐστίν] iki katları birbirlerine eşittirler]the doubles are equal to one another] ἴσον ἄρα ἐστὶ Eşittir dolaysıylaEqual therefore is καὶ τὸ ΒΛ παραλληλόγραμμον ΒΛ paralelkenarı daalso the parallelogram ΒΛ τῷ ΗΒ τετραγώνῳ ΗΒ karesineto the square ΗΒ ὁμοίως δὴ Şimdi benzer şekildeSimilarly then ἐπιζευγνυμένων τῶν ΑΕ ΒΚ birleştirildiğinde ΑΕ ve ΒΚthere being joined ΑΕ and ΒΚ δειχθήσεται goumlsterilecek kiit will be shown that καὶ τὸ ΓΛ παραλληλόγραμμον ΓΛ paralelkenarı daalso the parallelogram ΓΛ ἴσον τῷ ΘΓ τετραγώνῳ eşittir ΘΓ karesine[is] equal to the square ΘΓ ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον Dolayısıyla ΔΒΕΓ bir buumltuumln olarakTherefore the square ΔΒΕΓ as a δυσὶ τοῖς ΗΒ ΘΓ τετραγώνοις ΗΒ ve ΘΓ iki karesine

whole ἴσον ἐστίν eşittirto the two squares ΗΒ and ΘΓ καί ἐστι Ayrıcais equal τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ΒΔΕΓ karesi ΒΓ uumlzerine kurulmuşturAlso is ἀναγραφέν ve ΗΒ ve ΘΓ ΒΑ ve ΑΓ uumlzerinethe square ΒΔΕΓ set up on ΒΓ τὰ δὲ ΗΒ ΘΓ ἀπὸ τῶν ΒΑ ΑΓ Dolayısıyla ΒΓ kenarındaki kareand ΗΒ and ΘΓ on ΒΑ and ΑΓ τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγω- eşittirTherefore the square on the side ΒΓ νον ΒΑ ve ΑΓ kenarlarındaki karelereis equal ἴσον ἐστὶto the squares on the sides ΒΑ and τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-

ΑΓ τραγώνοις

Therefore in right-angled triangles ᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις Dolayısıyla dik accedilılı uumlccedilgenlerdethe square on the side subtending the τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτει- dik accedilının goumlrduumlğuuml kenar uumlzerindeki

right angle νούσης πλευρᾶς τετράγωνον kareis equal ἴσον ἐστὶ eşittirto the squares on the sides subtending τοῖς ἀπὸ τῶν τὴν ὀρθὴν [γωνίαν] περιε- dik accedilıyı iccedileren kenarların uumlzerindek-

the right [angle] χουσῶν πλευρῶν τετραγώνοις ileremdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Fitzpatrick considers this ordering of the two straight lines to belsquoobviously a mistakersquo But if it is a mistake how could it have beenmade

CHAPTER ELEMENTS

Β

Α

ΕΔ Λ

Γ

Ζ

Η

Θ

Κ

If of a triangle ᾿Εὰν τριγώνου Eğer bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining sides τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

of the triangle πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the two remaining sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktir

For of the triangle ΑΒΓ Τριγώνου γὰρ τοῦ ΑΒΓ Ccediluumlnkuuml ΑΒΓ uumlccedilgenininthe square on the one side ΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγω- bir ΒΓ kenarındaki karesimdashsuppose it is equal νον mdashvarsayılsın eşitto the squares on the sides ΒΑ and ἴσον ἔστω ΒΑ ve ΑΓ kenarlarındaki karelere

ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ πλευρῶν τε-τραγώνοις

I say that λέγω ὅτι İddia ediyorum kiright is the angle ΒΑΓ ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία ΒΑΓ accedilısı diktir

For suppose has been drawn ῎Ηχθω γὰρ Ccediluumlnkuuml ccedilizilmiş olsunfrom the point Α ἀπὸ τοῦ Α σημείου Α noktasındanto the straight ΑΓ τῇ ΑΓ εὐθείᾳ ΑΓ doğrusunaat rights πρὸς ὀρθὰς dik accedilılardaΑΔ ἡ ΑΔ ΑΔand let be laid down καὶ κείσθω ve yerleştirilmiş olsunequal to ΒΑ τῇ ΒΑ ἴση ΒΑ doğrusuna eşitΑΔ ἡ ΑΔ ΑΔand suppose ΔΓ has been joined καὶ ἐπεζεύχθω ἡ ΔΓ ve ΔΓ birleştirilmiş olsun

Since equal is ΔΑ to ΑΒ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ Eşit olduğundan ΔΑ ΑΒ kenarınaequal is ἴσον ἐστὶ eşittiralso the square on ΔΑ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον ΔΑ uumlzerindeki kare deto the square on ΑΒ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ ΑΒ uumlzerindeki kareyeLet be added in common κοινὸν προσκείσθω Eklenmiş olsun ortakthe square on ΑΓ τὸ ἀπὸ τῆς ΑΓ τετράγωνον ΑΓ uumlzerindeki karetherefore the squares on ΔΑ and ΑΓ τὰ ἄρα ἀπὸ τῶν ΔΑ ΑΓ τετράγωνα dolayısıyla ΔΑ ve ΑΓ uumlzerlerindekiare equal ἴσα ἐστὶ karelerto the squares on ΒΑ and ΑΓ τοῖς ἀπὸ τῶν ΒΑ ΑΓ τετραγώνοις eşittirBut those on ΔΑ and ΑΓ ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ ΑΓ ΒΑ ve ΑΓ uumlzerlerindeki karelereare equal ἴσον ἐστὶ Ama ΔΑ ve ΑΓ kenarları uumlzer-to that on ΔΓ τὸ ἀπὸ τῆς ΔΓ lerindekilerfor right is the angle ΔΑΓ ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία eşittirand those on ΒΑ and ΑΓ τοῖς δὲ ἀπὸ τῶν ΒΑ ΑΓ ΔΓ uumlzerlerindekineare equal ἴσον ἐστὶ ΔΑΓ accedilısı dik olduğundan

to that on ΒΓ τὸ ἀπὸ τῆς ΒΓ ve ΒΑ ile ΑΓ uumlzerlerindekilerfor it is supposed ὑπόκειται γάρ eşittirlertherefore the square on ΔΓ τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ΒΓ uumlzerlerindekineis equal ἴσον ἐστὶ ccediluumlnkuuml varsayıldıto the square on ΒΓ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ dolayısıyla ΔΓ uumlzerlerindekiso that the side ΔΓ ὥστε καὶ πλευρὰ eşittirto the side ΒΓ ἡ ΔΓ τῇ ΒΓ ΒΓ uumlzerlerindeki kareyeis equal ἐστιν ἴση boumlylece ΔΓ kenarıand since equal is ΔΑ to ΑΒ καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ ΒΓ kenarınaand common [is] ΑΓ κοινὴ δὲ ἡ ΑΓ eşittirthe two ΔΑ and ΑΓ δύο δὴ αἱ ΔΑ ΑΓ ve ΔΑ ΑΒ kenarına eşit olduğundanto the two ΒΑ and ΑΓ δύο ταῖς ΒΑ ΑΓ ve ΑΓ ortakare equal ἴσαι εἰσίν ΔΑ ve ΑΓ ikilisiand the base ΔΑ καὶ βάσις ἡ ΔΓ ΒΑ ve ΑΓ ikilisineto the base ΒΓ βάσει τῇ ΒΓ eşittirler[is] equal ἴση ve ΔΑ tabanıtherefore the angle ΔΑΓ γωνία ἄρα ἡ ὑπὸ ΔΑΓ ΒΓ tabanınato the angle ΒΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ eşittir[is] equal [ἐστιν] ἴση dolayısıyla ΔΑΓ accedilısıAnd right [is] ΔΑΓ ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ ΒΑΓ accedilısınaright therefore [is] ΒΑΓ ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ eşittir

Ve ΔΑΓ diktirdiktir dolayısıyla ΒΑΓ

If therefore of a triangle ᾿Εὰν ἀρὰ τριγώνου Eğer dolayısıyla bir uumlccedilgendethe square on one of the sides τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον bir kenarın uumlzerindeki karebe equal ἴσον ᾖ eşitseto the squares on the remaining two τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο uumlccedilgenin geriye kalan kenarlarındaki

sides πλευρῶν τετραγώνοις karelerethe angle contained ἡ περιεχομένη γωνία uumlccedilgenin geriye kalan kenarlarınca iccediler-by the remaining two sides of the tri- ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο ilen

angle πλευρῶν accedilıis right ὀρθή ἐστιν diktirmdashjust what it was necessary to show ὅπερ ἔδει δεῖξαι mdashgoumlsterilmesi gereken tam buydu

Α Β

Γ

Δ

Bibliography

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[] The thirteen books of Euclidrsquos Elements translated from the text of Heiberg Vol I Introduction andBooks I II Vol II Books IIIndashIX Vol III Books XndashXIII and Appendix Dover Publications Inc New York Translated with introduction and commentary by Thomas L Heath nd ed MR b

[] Euclidrsquos Elements Green Lion Press Santa Fe NM All thirteen books complete in one volumethe Thomas L Heath translation edited by Dana Densmore MR MR (j)

[] H W Fowler A dictionary of modern English usage second ed Oxford University Press revised andedited by Ernest Gowers

[] A dictionary of modern English usage Wordsworth Editions Ware Hertfordshire UK reprint ofthe original edition

[] Homer C House and Susan Emolyn Harman Descriptive english grammar second ed Prentice-Hall EnglewoodCliffs NJ USA Revised by Susan Emolyn Harman Twelfth printing

[] Rodney Huddleston and Geoffrey K Pullum The Cambridge grammar of the English language Cambridge Uni-versity Press Cambridge UK reprinted

[] Wilbur R Knorr The wrong text of Euclid on Heibergrsquos text and its alternatives Centaurus () no -ndash MR (c)

[] On Heibergrsquos Euclid Sci Context () no - ndash Intercultural transmission of scientificknowledge in the middle ages Graeco-Arabic-Latin (Berlin ) MR (b)

[] Henry George Liddell and Robert Scott A Greek-English lexicon Clarendon Press Oxford revised and aug-mented throughout by Sir Henry Stuart Jones with the assistance of Roderick McKenzie and with the cooperationof many scholars With a revised supplement

[] James Morwood and John Taylor (eds) The pocket Oxford classical Greek dictionary Oxford University PressOxford

[] Reviel Netz The shaping of deduction in Greek mathematics Ideas in Context vol Cambridge UniversityPress Cambridge A study in cognitive history MR MR (f)

[] Allardyce Nicoll (ed) Chapmanrsquos Homer The Iliad paperback ed Princeton University Press Princeton NewJersey With a new preface by Garry Wills Original publication

[] Proclus A commentary on the first book of Euclidrsquos Elements Princeton Paperbacks Princeton University PressPrinceton NJ Translated from the Greek and with an introduction and notes by Glenn R Morrow Reprintof the edition With a foreword by Ian Mueller MR MR (k)

[] Atilla Oumlzkırımlı Tuumlrk dili dil ve anlatım [the turkish language language and expression] İstanbul Bilgi Uumlniver-sitesi Yayınları Yaşayan Tuumlrkccedile Uumlzerine Bir Deneme [An Essay on Living Turkish]

[] Herbert Weir Smyth Greek grammar Harvard University Press Cambridge Massachussets Revised byGordon M Messing Eleventh Printing Original edition

[] Ivor Thomas (ed) Selections illustrating the history of Greek mathematics Vol II From Aristarchus to PappusHarvard University Press Cambridge Mass With an English translation by the editor MR b

[] Henry David Thoreau Walden [] and other writings Bantam Books New York Edited and with anintroduction by Joseph Wood Krutch