5 Hyperbolic Triangle Geometry - UNC Charlottemath2.uncc.edu/~frothe/3181alllhyp1_5.pdf5 Hyperbolic...

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5 Hyperbolic Triangle Geometry 5.1 Hyperbolic trigonometry Figure 5.1: The trigonometry of the right triangle. Theorem 5.1 (The hyperbolic right triangle). The sides and angles of any hyper- bolic right triangle with γ the right angle satisfy sin α = sinh a sinh c and sin β = sinh b sinh c (5.1) cos α = tanh b tanh c and cos β = sinh a sinh c (5.2) cosh c = cosh a cosh b (5.3) cosh a = cos α sin β and cosh b = cos β sin α (5.4) cosh c = cot α cot β (5.5) Proof of the formula (5.2) for the cos of an angle. The formula has to be derived from the geometry in the Poincar´ e and Klein model, which are best used together. 827

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5 Hyperbolic Triangle Geometry

5.1 Hyperbolic trigonometry

Figure 5.1: The trigonometry of the right triangle.

Theorem 5.1 (The hyperbolic right triangle). The sides and angles of any hyper-bolic right triangle with γ the right angle satisfy

sinα =sinh a

sinh cand sin β =

sinh b

sinh c(5.1)

cosα =tanh b

tanh cand cos β=

sinh a

sinh c(5.2)

cosh c = cosh a cosh b(5.3)

cosh a =cosα

sin βand cosh b =

cos β

sinα(5.4)

cosh c = cotα cot β(5.5)

Proof of the formula (5.2) for the cos of an angle. The formula has to be derived fromthe geometry in the Poincare and Klein model, which are best used together.

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We depict the given right triangle �ABC in the Poincare model with vertex A = Oat the center. The opposite side is depicted as a circle around a⊥. Let the transforma-tion to the Klein model produce triangle �ALK. Still, the right angle at vertex C istransformed to a right angle at K, since AK is a diameter.We use the underlying Euclidean geometry for the �ALK and obtain

cosα =|OK||OL| =

tanh b

tanh c

We have used that the hyperbolic distance to the center in the Klein model is given by

b = s(O,K) =1

2ln1 + |OK|1− |OK| = tanh

−1 |OK|

Hence |OK| = tanh b, and similarly |OL| = tanh c.Proof of the formula (5.1) for the sin of an angle. We use the same setting as in theproof given above. Once more, the given right triangle �ABC is depicted with vertexA = O at the center. In the Poincare disk model, the opposite side is depicted as acircle around a⊥ = K ′. The translation to Klein’s model produces the points C �→ Kand B �→ L, as well as for their images C ′ �→ K ′ and B′ �→ L′ by circular inversion. Theunderlying Euclidean geometry for the �K ′L′B yields

sin β =|BL′||BK ′| =

|BL′||CK ′| =

|BB′||CC ′|

We use now the hyperbolic distance to the center in the Poincare model:

b = s(O,C) = ln1 + |OC|1− |OC| = 2 tanh

−1 |OC|

Hence |OC| = tanh b2and

|CC ′| = |OC ′| − |OC| = sinh(b/2)cosh(b/2)

− cosh(b/2)sinh(b/2)

=2

sinh b

Similarly we get |BB′| = 2sinh c

. Thus the required sin is

sin β =|BB′||CC ′| =

sinh b

sinh c

All remaining formulas can be derived using only trigonometric identities.

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Proof of the Pythagorean formula (5.3). Because of the Pythagorean theorem in its Eu-clidean form, we know that

1 = cos2 α + sin2 α =tanh2 b

tanh2 c+sinh2 a

sinh2 csinh2 c · cosh2 b = sinh2 b · cosh2 c+ sinh2 a · cosh2 b

(1 + sinh2 c) · cosh2 b = sinh2 b · cosh2 c+ (1 + sinh2 a) · cosh2 bcosh2 c · cosh2 b− sinh2 b · cosh2 c = cosh2 a · cosh2 b

cosh2 c = cosh2 a · cosh2 bcosh c = cosh a · cosh b

A side in terms of two angles (5.4) . We get from the hyperbolic Pythagorean theorem

cosh a =cosh c

cosh b=tanh b

tanh c:sinh b

sinh c=cosα

sin β

Similarly, we get

cosh b =cos β

sinα

Formula (5.5) now follows immediately from the hyperbolic Pythagorean theorem (5.3).

Problem 5.1 (The hyperbolic area of circle, calculated in Archimedes fash-ion). We use a regular n-gon with radius of circum-circle R to get the area of the circlefrom the of the defect, taking the limit n → ∞.(a) We calculate cotα for a right triangle with hypothenuse R and β = π

n.

(b) From the defect δ = π2− α − β of one of 2n right triangles, we add up the total

defect for the n-gon to be

Δn = 2n[π2− α

]− 2π

(c) We need the first term in the expansions a0 +a1n+ a2

n2 + . . . for the quantities

tan(π2− α

)= cotα =(5.6)

π

2− α =(5.7)

Δn =(5.8)

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From the last expansion, we calculate limn→∞Δn, which is the area of the circle withradius R.

Answer. (a) The regular n-gon consists of n nonoverlapping isosceles triangles. Wedrop the perpendiculars from the center O and get 2n right triangles. Each oneof them has the angle β = π

nat the center and the hypothenuse R. From the

trigonometry of the right triangle we get

cotα cot β = coshR

cotα = (coshR) tanπ

n

(c) We need the first term in the expansions a0 +a1n+ a2

n2 + . . . for the quantities

tan(π2− α

)= cotα = (coshR) tan

π

n= (coshR)

π

n+O(n−3)

(5.9)

π

2− α = arctan

[(coshR)

π

n+O(n−3)

]= (coshR)

π

n+O(n−3)(5.10)

Δn = 2n[π2− α

]− 2π = 2π(coshR− 1) +O(n−2)

(5.11)

limn→∞

Δn = 2π(coshR− 1) = 4π sinh2 R2

(5.12)

which is the area of the circle with radius R.

5.2 The orthocenter

Theorem 5.2 (The conditional orthocenter—hyperbolic version). The threealtitudes of an acute or right triangle always intersect in one point. For an obtusetriangle, the three altitudes may intersect or not. There are three possible cases:

(i) The three altitudes of an obtuse triangle intersect in one point.

(ii) The three altitudes of an obtuse triangle are all divergent parallel to each other.There exists a line p perpendicular to all three altitudes.

(iii) The three altitudes of an obtuse triangle are all asymptotically parallel to eachother.

Corollary 60. 60 If any two altitudes intersect, then all three altitudes intersect in onepoint.

60The first two sentences of the Corollary are stated in neutral geometry. A proof of that part inneutral geometry is clearly valid a real—but not Klein-bottle—of wine.

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If the altitudes of two sides of a triangle have a common perpendicular, then thealtitudes of all three sides have a common perpendicular.If the altitudes of two sides of a triangle are asymptotically parallel, then the altitudes

of all three sides are asymptotically parallel.

Figure 5.2: If any two altitudes intersect, one can put the intersection point at the centerof Klein’s disk. Three drawings are given: for an acute, right, or obtuse triangle.

We need to clarify some terms about the use of any mathematical models, as Klein’sor Poincare’s:

Definition 5.1. A theorem or a feature of a figure is part of neutral geometry if andonly if it can be deducted assuming only the axioms of incidence, order, congruence.The facts of neutral geometry are valid in both Euclidean and hyperbolic geometry—

as well as the more exotic non-Archimedean geometries.

Definition 5.2. A feature of a figure drawn inside Klein’s model (as for example anangle, midpoint, altitude or bisector) is called absolute if it appears in the same way bothfor the underlying Euclidean plane, on which the model is based, and the hyperbolicgeometry inside the model.

Remark. Here are some features that appear absolute, both as features of hyperbolicgeometry and in the underlying Euclidean plane: An angle with the center of Klein’sdisk appears as an absolute angle. A right angle of which one side is a diameter appearsabsolute. A perpendicular bisector or an angle bisector which is a diameter appearsabsolute.

Proof, using Klein’s model of hyperbolic geometry. As stated in Proposition 9.5, any twoaltitudes of an acute or right triangle do intersect. It can, but does not need to hap-pen that the altitudes of an obtuse triangle intersect. These are simple facts in neutralgeometry.If any two altitudes of the triangle intersect, we obtain an easily understandable

picture in Klein’s model: just put the intersection point H into the center of Klein’s

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disk. In the figure on page 831, I have given illustrations for an acute, right and anobtuse triangle.Why does that picture come out that simple? Here are the logical steps for the

reason: The right angles at the foot points of the two intersecting altitudes appearundistorted as absolute right angles. We can now apply Theorem 8.3 from Euclideangeometry to the Euclidean triangle in the underlying Euclidean plane of Klein’s model.Hence the third Euclidean altitude passes through the intersection point of the two otherones, which we have put at the center O. Because the right angle at the foot point ofthe third altitude is depicted undistorted, too, all three altitudes are both Euclidean aswell hyperbolic altitudes. Hence the three hyperbolic altitudes intersect at the centerH.For an obtuse triangle, two new cases (ii) and (iii) do occur in hyperbolic geometry.

They are illustrated in the figure on page 832. Of course, one can no longer put theorthocenter in the center of Klein’s disk, because it does not exist! Instead, we put the

Figure 5.3: The altitudes of an obtuse triangle may or may not intersect. A convenientdrawing puts the vertex with the obtuse angle at the center of Klein’s disk.

vertex C with the obtuse angle into the center of the disk. In that way, still the threeright angles at the foot points of the altitudes are depicted as absolute right angles. Allthree altitudes are both Euclidean as well as hyperbolic altitudes. Depending where thealtitudes intersect, we get the three cases:

(i) The intersection point H lies inside Klein’s disk. The three altitudes intersect inone point.

(ii) The intersection point H lies outside Klein’s disk. The three altitudes are alldivergent parallel to each other. There exists a line l perpendicular to all threealtitudes.

(iii) The intersection point H lies on the boundary of Klein’s disk. The three altitudesare all asymptotically parallel to each other.

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In the ”genuine hyperbolic case” (ii), point H lies outside Klein’s disk. Hence H is nota point of the hyperbolic plane, but a so-called ultra-ideal point. The polar of point Hyields a common perpendicular l to the three altitudes.Recall that the boundary circle δD of Klein’s disk is called the circle of infinity. Fi-

nally, we get the ”borderline case” (iii) if point H lies on the circle of infinity. Neithertwo of the three altitudes intersect, nor do any two of them have a common perpendic-ular.

5.3 About the circum-circle

We begin by recalling Propositions 9.2 and 9.3 from neutral triangle geometry.

Theorem 5.3 (The perpendicular bisectors and their meaning). Given anytriangle, three different cases can occur.

”As seen in Euclidean geometry” The three perpendicular bisectors intersect in onepoint. The triangle has a circum-circle, and the intersection point is the center ofthe circum-circle. This case always occurs for an acute or right midpoint-triangle.Too, it can—but does not need to occur—if the midpoint-triangle is obtuse.

The orthocenter of the midpoint triangle �MaMbMc, is the circum-center of thelarger original triangle �ABC, too.

(H2O) H2 = O

”The genuine hyperbolic case” The three bisectors are all divergently parallel toeach other. There exists a line l perpendicular to all three bisectors. All threevertices have the same distance from line l. Hence there exists an equidistanceline through the three vertices of the triangle.

”The borderline case” The three perpendicular bisectors are all asymptotically par-allel to each other. Neither two of the three bisectors intersect, nor do any twoof them have a common perpendicular. The three vertices lie neither on a circlenor an equidistance line. There exists a horocycle through the three vertices of thetriangle.

In hyperbolic geometry, all three cases do occur, whereas Euclidean geometry leads alwaysto the first case.

Corollary 61. Any two perpendicular bisectors of sides of a triangle intersect iff thebisectors of all three sides intersect in one point iff the triangle has a circum-circle.The perpendicular bisectors of any two sides of a triangle have a common perpen-

dicular iff the perpendicular bisectors of all three sides have a common perpendicular iffthe three vertices lie on an equidistance line.The perpendicular bisectors of two sides of a triangle are asymptotically parallel iff the

perpendicular bisectors of all three sides are asymptotically parallel iff the three verticeslie on a horocycle.

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Figure 5.4: The bisectors and altitudes of the midpoint triangle inside Klein’s disk model.For a triangle with circum-center, three drawings are given: for an acute, right, or obtusemidpoint triangle.

Proof for hyperbolic geometry, using Klein’s model. Most statements just repeat Propo-sition 9.6 from neutral geometry. As a new feature in hyperbolic geometry, we can useKlein’s model. Thus we can confirm that all three mentioned cases actually do occur.Too, the meaning of the ”borderline case” is clarified.By Corollary 24, the bisectors of the sides of the original triangle �ABC are the

altitudes of the midpoint-triangle�MaMbMc. Thus they form six right angles, in neutralgeometry.We start with the case that any two altitudes of the midpoint-triangle do intersect.

By Corollary 24, the altitudes of the midpoint-triangle are side bisectors for the origi-nal triangle. Now Proposition 9.2 implies that all three side bisectors intersect in onepoint. This point is the circum-center of the original triangle and the orthocenter of themidpoint-triangle as stated in formula (H2O). We have arrived at the case ”As seen inEuclidean geometry”, for which the triangle has a circum-circle.The case ”As seen in Euclidean geometry” always occurs for an acute or right

midpoint-triangle, because any two altitudes of an acute or right triangle do inter-sect by Proposition 9.5. It can, but does not need to happen that the altitudes of anobtuse midpoint-triangle intersect. In Klein’s model an easily understandable picture isobtained by putting point O into the center of Klein’s disk. In the figure on page 834,I have given illustrations for an acute, right and an obtuse midpoint-triangle.

Remark. For an acute midpoint-triangle, both vertex C and circum-center O lie on thesame side of the longer triangle side AB. For an obtuse midpoint-triangle, vertex C andcircum-center O lie opposite sides of the longer triangle side AB.

But, indeed, the other two cases already mentioned in Proposition 9.6 do occur inhyperbolic geometry, too. They are illustrated in the figure on page 835. Of course,one can no longer put the circum-center in the center of Klein’s disk, because it doesnot exist! Instead, we put the midpoint Mc of the longest side in the center of the

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Figure 5.5: The altitudes of the midpoint triangle intersect outside Klein’s disk.

disk. Of the six right angles between the perpendicular bisectors and the sides of thetriangles �ABC and �MaMbMc, only four are depicted undistorted. Because the threeright angles between the bisectors and the sides of the midpoint-triangle �MaMbMc arestill undistorted, it is straightforward to construct its virtual orthocenter H2. In the”genuine hyperbolic case”, point H2 lies outside Klein’s disk.Hence H2 is not a point of the hyperbolic plane. Instead, the polar of the ultra-ideal

point H2 yields a common perpendicular l to the three altitudes of triangle �MaMbMc.But—since Proposition 24 is a theorem of neutral geometry—it is still true that the

altitudes of the midpoint-triangle are the bisectors of the original triangle. Thereforeline l is a common perpendicular to the three bisectors of triangle �ABC.Now still we drop the perpendiculars from the three vertices A,B and C onto line l.

We end up with six lines that form right angles with line l! Too, we have reconstructedthe situation from Example 9.1. The three points X, Y and Z are the intersectionsof l with the perpendicular bisectors. Once more, we get three Saccheri quadrilateral�Y XAB and �Y ZCB and �ZXAC. Hence the three vertices A,B and C have con-gruent distances to line l, as to be shown.Recall that the boundary circle δD of Klein’s disk is called the circle of infinity.

Finally, we get the ”borderline case” if point H2 lies on the circle of infinity. In thatcase, the three vertices lie neither on a circle nor an equidistance line. Neither two ofthe three bisectors intersect, nor do any two of them have a common perpendicular.

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Figure 5.6: The three vertices A, B and C have congruent distances from the baseline l,which is chosen to be the horizontal diameter. The second drawing erases the part of theconstruction done outside the Poincare disk.

Problem 5.2. In the case that the three vertices of a triangle lie on an equidistanceline, they have congruent distances to the baseline. In the figure of page 835, we getAX ∼= BY ∼= CY , where X, Y, Z are the foot points of the perpendiculars, dropped fromthe vertices onto the baseline l. This is hard to believe, especially because segment CZintersects the opposite side AB of the triangle �ABC, but the other two segments AXand BY do not intersect any side of the triangle.Redraw the figure in Poincare’s model. The visible divergence of parallel lines makes

existence of a baseline l more intuitive.

Problem 5.3. In Klein’s disk model, there is given a triangle �ABC with midpointMc of side AB at the center. Construct the midpoint-triangle, the six right angles anddecide which case of Theorem 5.3 did occur.

Problem 5.4. Given given is a triangle �ABC with midpoint Mc of side AB at thecenter of a Klein disk, and midpoint Mb already specified. Construct the circle of infinityδD of Klein’s disk. Finally get the midpoint-triangle, the six right angles and decidewhich case of Theorem 5.3 did occur.

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Figure 5.7: Construction of the perpendicular bisector pb and the point H2 = O.

Figure 5.8: Construction of the circle of infinity δD and the point H2 = O.

5.4 Thales’ Theorem in hyperbolic geometry

As already mentioned in the introductory section, Thales’ theorem does not hold inhyperbolic geometry. Once it is known that the angle sum of a triangle is less than tworight angles, it is easy to see that the angle in a semicircle is acute. Nevertheless, evenin neutral geometry, we can get a nice statement about a triangle in a semicircle.

Proposition 5.1. Let one side of a triangle be a diameter of a semicircle and the third

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Figure 5.9: Construction of the circle of infinity δD and the point H2 = O—in this exampleit turns out to be outside the disk!

vertex lie on that semicircle. Then the third vertex and the midpoints of the three sidesare a Lambert quadrilateral.

Figure 5.10: A triangle in a semicircle produces a Lambert quadrilateral. For the proof,the extra line we need is—once again—the radius to the third vertex.

Proof. Let AB be the diameter and let C be the third vertex on the semicircle. Let�MaMbMc be the midpoint-triangle. BecauseMc has congruent distances from vertices

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A and C, it lies on the bisector pb, as follows from Proposition 9.1. Thus angle ∠McMbCis a right angle.By the same reasoning, congruent distances McB ∼= McC imply that Mc lies on the

bisector pa, and hence angle ∠McMaC is right.The bisector pb bisects the angle ∠AMcC, and the bisector pa bisects the supplemen-

tary angle ∠BMcC. As shown in paragraph of in-circle and ex-circles, the bisectors ofsupplementary angles are perpendicular. Hence pa and pb are perpendicular, and henceangle ∠MaMcMb is right.Thus we have checked that the quadrilateral �CMaMcMb has three right angles,

resulting in a Lambert quadrilateral.

Corollary 62. In neutral geometry, the angle in a semicircle is a right angle if andonly if a rectangle exists.

Reason. If a rectangle exists, we know from the second Legendre theorem that everyLambert quadrilateral is a rectangle. Hence the Lambert quadrilateral constructed inProposition 5.1 is a rectangle. The right angle at vertex C is now Thales’ right angle ina semicircle.Conversely, if the angle in a semicircle is right, the Lambert quadrilateral constructed

in Proposition 5.1 is a rectangle, and hence a rectangle does exist.

Remark. The corollary holds without assuming Archimedes’ axiom.

Similarly as we did in the introductory section, one can ask what happens for thethird vertex inside or outside the semicircle. Thus one is lead to a strengthened version,and a converse to Proposition 5.1.

Proposition 5.2 (A neutral version of the strengthened Thales’ theorem). Let one sideAB of a triangle be a diameter of a circle. If the third vertex C lies inside the circle, themidpoint-triangle has an obtuse angle at the center of the semicircle. The quadrilateral�CMaMcMb has an obtuse angle at vertex Mc and two acute angles at vertices Ma andMb.If the third vertex C lies outside the circle, the midpoint-triangle has an acute angle

at the center of the semicircle. The quadrilateral �CMaMcMb has an acute angle atvertex Mc and two obtuse angles at vertices Ma and Mb.

Corollary 63. Let one side of a triangle be a diameter of a circle. The third vertex Clies on that circle if and only if the midpoint-triangle has a right angle at the center ofthe semicircle.

Proof in Klein’s model of hyperbolic geometry. The drawings on page 840 use Klein’smodel with midpoint Mc at the center of the disk. I did not draw the circle of infinity∂D. The drawings include the semicircle, the original triangle �ABC, its midpoint-triangle �MaMbMc. By Corollary 24, the altitudes of the midpoint triangle are the side

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Figure 5.11: A triangle with third vertex inside the circle produces an obtuse angle of themidpoint-triangle at center Mc. In case of the third vertex outside the circle, an acute angleis produced.

bisectors of the original triangle. Hence the altitude of the midpoint-triangle droppedfrom vertex Mb onto the opposite side MaMc is equal to the perpendicular bisector pb.It has the foot point F . An additional perpendicular is dropped from Mc onto the sideAC, and has the foot point G. Begin by distinguishing the following cases:

(i) The midpoint-triangle has an obtuse angle at vertex Mc.

(ii) The midpoint-triangle has a right angle at vertex Mc.

(iii) The midpoint-triangle has an acute angle at vertex Mc, but an obtuse or rightangle at vertex Ma.

(iv) The midpoint-triangle has an acute angle at vertex Mc, but an obtuse or rightangle at vertex Mb.

(v) The midpoint-triangle is acute.

To finish the proof of the Corollary, assume that case (ii) occurs. Clearly the ”water-point” is H2 = O = F = Mc. The side bisector pb = MbFH2 and the line GMc areequal, because both are the perpendicular toMcMa at pointMc = F . We get a Lambertquadrilateral �CMaMcMb. Because Mc lies on the bisector pb, both points C and Ahave congruent distances to Mc. Hence point C lies on the circle with diameter AB.Next we consider case (i). By Proposition 9.5, the orthocenter H2 = O of the

midpoint-triangle lies inside the vertical angle to the obtuse angle ∠MbMcMa.61 Hence

H2 lies on the side of line AB opposite to the points Ma,Mb and C.

61Clearly, the orthocenter can be an ideal or ultra-ideal point of Klein’s model.

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Answer. Because the angle ∠MaMcMb is obtuse, the foot point F lies on the extensionof side MaMc of the midpoint-triangle, and hence Mc lies between F and Ma. Sortingagain points below and above line AB, we conclude that F and H2 lie below line AB,whereas Ma,Mb and C lie above line AB.

From here, one can check that ray−−−→MbMc lies inside the right angle ∠CMbH2. Hence

G lies on the ray−−→MbC, and GC is shorter than AC. Finally McC is shorter than McA,

which confirms that C lies inside the circle with diameter AB.It is left to the reader to check that in all remaining cases (iii) through (v) the

following can be deducted: Ray−−−→MbMc lies inside the right angle ∠AMbH2. Hence GC

is longer than AC. Finally McC is longer than McA, which confirms that C lies outsidethe circle with diameter AB.

Problem 5.5. Given is a triangle �ABC with midpoint Mc of longest side AB at thecenter of a Klein disk and vertex C in the upper half disk. Use Proposition 5.2, and provethat the hyperbolic midpoints Mb and Ma are below the apparent Euclidean midpoint ifand only if the midpoint-triangle is obtuse. The hyperbolic midpoints Mb and Ma areabove the apparent Euclidean midpoint if and only if the midpoint-triangle is acute.

Partial solution. The hyperbolic midpoint Mb is equal to the apparent midpoint of seg-ment AC if and only if Mb = G. By the neutral version of the strengthened Thales’theorem, given in Proposition 5.2 above, this happens if and only if the angleMaMcMb isright. The assertion follows by continuity, once is hold in at least one example for obtuseand acute triangles. This can be established for some extreme isosceles triangles.

5.5 The centroid

Theorem 5.4 (The Centroid). The three medians of a triangle intersect in one point.

Remark. The ratio in which the intersection point divides the medians can be differentfrom 2 : 1.

Proof. I use Klein’s model of hyperbolic geometry, to given a proof for the hyperboliccase. Given is a triangle �ABC. A convenient position is achieved by putting triangleside AB on the horizontal diameter with its midpoint Mc = O in the center of the disk.The midpoints of the two other sides of the triangle do not appear as absolute

midpoints, because of the distortion of distances in Klein’s model. Nevertheless, wecan take advantage of Theorem 9.1. The perpendicular bisector pc of one side AB isperpendicular to the line l through the midpoints Ma and Mb of the other two sides BCand CA. Because this perpendicular bisector pc is a vertical diameter of the disk D, theright angle between pc and l appears as an absolute right angle.Hence, in the sense of underlying Euclidean geometry, the lines AB and MaMb get

parallel . We are just back to the figure studied in the construction 14.3 of a Euclidean

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Figure 5.12: The three medians intersect in one point because the harmonic quadrilateral�ABMbMa has Euclidean parallel sides.

parallel— in the section about constructions by restricted means. By the Theorem ofthe harmonic quadrilateral (see Theorem 14.1), the three lines CMc, AMa and BMb

intersect in one point, as to be shown.

5.6 The in-circle, ex-circles, and the orthic triangle

Problem 5.6 (Conditional ex-circle—hyperbolic version). Provide the details of theproof for the following proposition. Part is easy to verify in Klein’s model—once oneknows how to interpretate Klein’s model.If any two of the exterior bisectors at vertices A and B, and the interior bisector of

the third angle ∠BCA intersect, then all these three bisectors intersect in one point. Inthat case, the triangle has an ex-circle touching side AB from outside, and the extensionsof the two other sides.If any two of these three bisectors have a common perpendicular, then three have a

common perpendicular. Furthermore, there exists a unique equidistance line touchingside AB from outside, and the extensions of the two other sides.If any two of these three bisectors are asymptotically parallel, then all three are

asymptotically parallel. Furthermore, there exists a unique horocycle touching side ABfrom outside, and the extensions of the two other sides.

Theorem 5.5. For an acute triangle, the altitudes are the inner angular bisectors ofthe orthic triangle.For an obtuse triangle, only the altitude dropped from the obtuse angle is an inner

angular bisector. The other two altitudes are exterior angular bisectors of the orthictriangle.

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Figure 5.13: The altitudes are angular bisectors of the orthic triangle.

Proof. This is easy to verify in Klein’s model! One needs to choose the center of thedisk differently for an acute and an obtuse triangle. For an acute triangle, I put theorthocenter H at the center of the disk. Again the altitudes as well as the angularbisectors and the in-circle of the orthic triangle become absolute.For an obtuse triangle, I put vertex C at the center of the disk. Still the right angles

at the foot points of the three altitudes appear as absolute right angles. But only thealtitude dropped from the obtuse angle is an inner angular bisector of the orthic triangle.The other two inner bisectors are just the extensions of the two sides of the triangle atthe obtuse angle. The other two altitudes get exterior angular bisectors, since it is afact of neutral geometry that interior an exterior angular bisectors are perpendicular toeach other.

Problem 5.7. Prove that the orthic triangle of an acute triangle has an in-circle andthree ex-circles.The orthic triangle of an obtuse triangle has an in-circle and either two or three

ex-circles. The orthic triangle has three ex-circles if and only if the original triangle hasan orthocenter.

Proof. For an acute triangle, one can put any vertex at the center of Klein’s disk,too. One sees that the center becomes the intersection of two exterior and one interiorangular bisector, and one gets an ex-circle around that center, all depicted in Klein’smodel absolutely. (Of course, one can only depict one of the four circles in questionabsolutely at a time!).

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Figure 5.14: Centers for any ex-circle of the orthic triangle can only be the three verticesor the orthocenter of the original triangle.

For an obtuse triangle, one can put the two vertices with acute angles at the center,and get two ex-circles of the orthic triangle. To get the third ex-circle, one needs toput the orthocenter at the center of Klein’s disk, which is possible if and only if theorthocenter exists.

Figure 5.15: The triangle has no circum-circle—drawn in Klein’s model!

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