Congruent Polygons and Congruent Parts

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1 Congruent Polygons and Congruent Parts Two polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent Corresponding parts of congruent polygons are congruent. Congruent Triangles The following also holds true for Triangles: Reflexive Property ΔABC ΔABC Symmetric Property If ΔABC ΔDEF then ΔABC ΔDEF Transitive Property If ΔABC ΔDEF and ΔDEF ΔRST then ΔABC ΔRST Proving Triangles Congruent Side Angle Side Two triangles are congruent if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of the other. SAS SAS Angle Side Angle Two triangles are congruent if two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the other. ASA ASA Side Side Side Two triangles are congruent if three sides of one triangle are congruent respectively, to three sides of the other. SSS SSS Congruence Based on Triangles An Altitude of a triangle is a line segment drawn from any vertex of the triangle perpendicular to and ending in the line contains the opposite side. A Median of a triangle is a line segment that joins any vertex of the triangle to the midpoint of the opposite side. An Angle Bisector of a triangle is a line segment that bisects any angle of the triangle and terminates in the side opposite that angles. Using congruent triangles to prove line segments congruent and angles congruent. (Corresponding parts of congruent triangles are congruent)(CPCTC)

Transcript of Congruent Polygons and Congruent Parts

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Congruent Polygons and Congruent Parts Two polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent Corresponding parts of congruent polygons are congruent.

Congruent Triangles The following also holds true for Triangles: Reflexive Property ΔABC ≅ ΔABC Symmetric Property If ΔABC ≅ ΔDEF then ΔABC ≅ ΔDEF Transitive Property If ΔABC ≅ ΔDEF and ΔDEF ≅ ΔRST then ΔABC ≅ ΔRST

Proving Triangles Congruent

Side Angle Side Two triangles are congruent if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of the other.

SAS ≅ SAS

Angle Side Angle Two triangles are congruent if two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the other.

ASA ≅ ASA

Side Side Side Two triangles are congruent if three sides of one triangle are congruent respectively, to three sides of the other.

SSS ≅ SSS

Congruence Based on Triangles An Altitude of a triangle is a line segment drawn from any vertex of the triangle perpendicular to and ending in the line contains the opposite side. A Median of a triangle is a line segment that joins any vertex of the triangle to the midpoint of the opposite side.

An Angle Bisector of a triangle is a line segment that bisects any angle of the triangle and terminates in the side opposite that angles. Using congruent triangles to prove line segments congruent and angles congruent. (Corresponding parts of congruent triangles are congruent)(CPCTC)

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Isosceles Triangles

If two sides of a triangle are congruent, the angles opposite these sides are congruent. The median from the vertex angle of an isosceles triangle bisects the vertex angle. The median from the vertex angle of an isosceles triangle is perpendicular to the base.

Equilateral Triangle Every equilateral triangle is equiangular.

Perpendicular Bisector The Perpendicular Bisector of a line segment is any line or subset of a line that is perpendicular to the line at its midpoint. If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment If a point is equidistant from the endpoints of a line segment, then it is on the perpendicular bisector of the line segment. If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the endpoints of a line segment. The perpendicular bisectors of the sides of a triangle are congruent.

Methods of Proving Lines or Line Segments Perpendicular 1. The two lines form right angles at their point of intersection.

2. The two lines form congruent adjacent angles at their point of intersection.

3. Each of two points on one line is equidistant from the endpoints of a segment of the other.

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#1

Given: ΔABC CD bisects AB CD ⊥ AB Prove: ΔACD ≅ ΔBCD

Statement 1. ΔABC CD bisects AB CD ⊥ AB

Reasons 1. Given

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#2

Given: ABC and DBE bisect each

other.

Prove: ΔABD ≅ ΔCBD

Statement

Reasons

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#3

Given: AB = CD and BC = AD ∠DAB, ∠ABC, ∠BCD and ∠CDA

are rt ∠ Prove: ΔABC ≅ ΔADC

Statement

Reasons

#4

Given: ∠PQR ≅ ∠RQS PQ ≅ QS Prove: ΔPQR ≅ ΔRQS

Statement

Reasons

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#5

Given: AEB & CED intersect at E E is the midpoint AEB AC ⊥ AE & BD ⊥ BE Prove: ΔAEC ≅ ΔBED

Statement

Reasons

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#6

Given: AEB bisects CED AC ⊥ CED & BD ⊥ CED Prove: ΔEAC ≅ ΔEBD

Statement

Reasons

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#7

Given: ΔABC is equilateral D midpoint of AB Prove: ΔACD ≅ ΔBCD

Statement

Reasons

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#8

Given: m∠A = 50, m∠B = 45, AB = 10cm, m∠D = 50 m∠E = 45 and DE = 10cm Prove: ΔABC ≅ ΔDEF

Statement

Reasons

#9

Given: GEH bisects DEF m∠D = m∠F Prove: ΔGFE ≅ ΔDEH

Statement

Reasons

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#10

Given: PQ bisects RS at M ∠R ≅ ∠S Prove: ΔRMQ ≅ ΔSMP

Statement

Reasons

#11

Given: DE ≅ DG EF ≅ GF Prove: ΔDEF ≅ ΔDFG

Statement

Reasons

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#12

Given: KM bisects ∠LKJ LK ≅ JK Prove: ΔJKM ≅ ΔLKM

Statement

Reasons

#13

Given: . PR ≅ QR ∠P ≅ ∠Q RS is a median Prove: ΔPSR ≅ ΔQSR

Statement

Reasons

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#14

Given: EG is ∠ bisector EG is an altitude Prove: ΔDEG ≅ ΔGEF

Statement

Reasons

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#15

Given: ∠A and ∠D are a rt ∠ AE ≅ DF AB ≅ CD Prove: EC ≅ FB

Statement

Reasons

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#16

Given: AC ≅ BC D midpoint of AB Prove: ∠A ≅ ∠B

Statement

Reasons

#17

Given: . AB ≅ CD ∠CAB ≅ ∠ACD Prove: AD ≅ BC

Statement

Reasons

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#18

Given: AEB & CED bisect each

Other Prove: ∠C ≅ ∠D

Statement

Reasons

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#19

Given: ∠KLM & ∠NML are rt ∠ KL ≅ NM Prove: ∠K ≅ ∠N

Statement

Reasons

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#20

Given: AB ≅ BC ≅ CD PA ≅ PD & PB ≅ PC Prove: a) ∠APB ≅ ∠DPC b) ∠APC ≅ ∠DPB

Statement

Reasons

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#21

Given: PM is Altitude PM is median Prove: a) ΔLNP is isosceles b) PM is ∠ bisector

Statement

Reasons

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#22

Given: AC ≅ BC Prove: ∠CAD ≅ ∠CBE

Statement

Reasons

#23

Given: AB ≅ BC & AD ≅ CD Prove: ∠BAD ≅ ∠BCD

Statement

Reasons

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#24

Given: ΔABC ≅ ΔDEF M is midpoint of AB N is midpoint DE Prove: ΔAMC ≅ ΔDNF

Statement

Reasons

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#25

Given: ΔABC ≅ ΔDEF CG bisects ∠ACB FH bisects ∠DFE Prove: CG ≅ FH

Statement

Reasons

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#26

Given: ΔAME ≅ ΔBMF DE ≅ CF Prove: AD ≅ BC

Statement

Reasons

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#27

Given: AEC & DEB bisect each other Prove: E is midpoint of FEG

Statement

Reasons

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#28

Given: BC ≅ BA BD bisects ∠CBA Prove: BD bisects ∠CDA

Statement

Reasons

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#29

Given: AE ≅ FB AD ≅ BC ∠A and ∠B are Rt. ∠ Prove: ΔADF ≅ ΔCBE DF ≅ CE

Statement

Reasons

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#30

Given: SPR ≅ SQT PR ≅ QT Prove: ΔSRQ ≅ ΔSTP ∠R ≅ ∠T

Statement

Reasons

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#31

Given: AD ≅ BC AD ⊥ AB & BC ⊥ AB Prove: ΔDAB ≅ ΔCBA

AC ≅ BD

Statement

Reasons

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#32

Given: ∠BAE ≅ ∠CBF ∠BCE ≅ ∠CDF AB ≅ CD Prove: AE ≅ BF ∠E ≅ ∠F

Statement

Reasons

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#33

Given: TM ≅ TN M is midpoint TR N is midpoint TS Prove: RN ≅ SM

Statement

Reasons

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#34

Given: AD ≅ CE & BD ≅ EB Prove: ∠ADC ≅ ∠CEA

Statement

Reasons

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#35

Given: AE ≅ BF & AB ≅ CD ∠ABF is the suppl. of ∠A Prove: ΔAEC ≅ ΔBFD

Statement

Reasons

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#36

Given: AB ≅ BC BD bisects ∠ABC Prove: AE ≅ CE

Statement

Reasons

#37

Given: PB ≅ PC Prove: ∠ABP ≅ ∠DCP

Statement

Reasons

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#38

Given: AC and BD are ⊥ bisectors of each other. Prove: AB ≅ BC ≅ CD ≅ AD

Statement

Reasons

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#39

Given: AEFB, ∠1 ≅ ∠2 CE ≅ DF , AE ≅ BF Prove: ΔAFD ≅ ΔBEC

Statement

Reasons

#40

Given: SX ≅ SY , XR ≅ YT Prove: ΔRSY ≅ ΔTSX

Statement

Reasons

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#41

Given: AD ≅ BC AD ⊥ AB , BC ⊥ AB Prove: ΔDAB ≅ ΔCBA

Statement

Reasons

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#42

Given: AF ≅ CE ∠1 ≅ ∠2, ∠3 ≅ ∠4 Prove: ΔABE ≅ ΔCDF

Statement

Reasons

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#43

Given: AB ⊥ BF , CD ⊥ BF ∠1 ≅ ∠2, BD ≅ FE Prove: ΔABE ≅ ΔCDF

Statement

Reasons

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#44

Given: ∠BAC ≅ ∠BCA CD bisects ∠BCA AE bisects ∠BAC Prove: ΔADC ≅ ΔCEA

Statement

Reasons

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#45

Given: TR ≅ TS , MR ≅ NS Prove: ΔRTN ≅ ΔSTM

Statement

Reasons

#46

Given: CEA ≅ CDB, ΔABC AD and BE intersect at P ∠PAB ≅ ∠PBA Prove: PE ≅ PD

Statement 1.

Reasons

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#47

Given: AB ≅ AD and BC ≅ CD Prove: ∠1 ≅ ∠2

Statement

Reasons

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#48

Given: BD is both median and

altitude to AC Prove: BA ≅ BC

Statement

Reasons

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#49

Given: ∠CDE ≅ ∠CED and AD ≅ EB Prove: ∠ACD ≅ ∠BCE

Statement

Reasons

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#50

Given: Isosceles triangle CAT CT ≅ AT and ST bisects ∠CTA Prove: ∠SCA ≅ ∠SAC

Statement

Reasons

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#51

Given: ∠1 ≅ ∠2 BD ⊥ AC Prove: ΔABD ≅ ΔCBD

Statement

Reasons

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#52

Given: ∠P ≅ ∠S R is midpoint of PS Given: ΔPQR ≅ ΔSTR

Statement

Reasons

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#53

Given: FG ⊥ DE G is midpoint of DE Given: ΔDFG ≅ ΔEFG

Statement

Reasons

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#54

Given: AC ≅ BC D is midpoint of AB Prove: ΔACD ≅ ΔBCD

Statement

Reasons

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#55

Given: PT bisects QS PQ ⊥ QS and TS ⊥ QS Prove: ΔPQR ≅ ΔRST

Statement

Reasons

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#56

Given: AB ≅ DE and FE ≅ BC FE ⊥ AD and BC ⊥ AD Prove: ΔAEF ≅ ΔCBD

Statement

Reasons

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#57

Given: SM is ⊥ bisector of LP RM ≅ MQ ∠a ≅ ∠b Prove: ΔRLM ≅ ΔQPM

Statement

Reasons

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#59

Given: AC ≅ BC CD ⊥ AB Prove: ΔACD ≅ ΔBCD

Statement

Reasons

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#60

Given: FQ bisects AS ∠A ≅ ∠S Prove: ΔFAT ≅ ΔQST

Statement

Reasons

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#61

Given: ∠A ≅ ∠D and ∠BCA ≅ ∠FED AE ≅ CD ∠AEF ≅ ∠BCD Prove: ΔABC ≅ ΔDFE

Statement

Reasons

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#62

Given: SU ≅ QR , PS ≅ RT ∠TSU ≅ ∠QRP Prove: ΔPQR ≅ ΔSTU ∠Q ≅ ∠U

Statement

Reasons

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#63

Given: ∠M ≅ ∠D ME ≅ HD ∠THE ≅ ∠SEM Prove: ΔMTH ≅ ΔDSE

Statement

Reasons

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#64

Given; SQ bisects ∠PSR ∠P ≅ ∠R Prove: ΔPQS ≅ ΔQSR

Statement

Reasons

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#65

Given: PQ ⊥ SQ and TS ⊥ QS R midpoint of QS Prove: ∠P ≅ ∠T

Statement

Reasons

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#66

Given: BC ≅ FB , BT ≅ BV DV ≅ TS , CD ≅ FS Prove: ∠D ≅ ∠S

Statement

Reasons

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#67

Given: PQ ≅ DE and PB ≅ AE QA ⊥ PE and BD ⊥ PE Prove: ∠D ≅ ∠Q

Statement

Reasons

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#68

Given: TS ≅ TR ∠P ≅ ∠Q Prove: PS ≅ QR

Statement

Reasons

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#69

Given: HY and EV bisect each other Prove: HE ≅ VY

Statement

Reasons

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#70

Given: ∠E ≅ ∠D and ∠A ≅ ∠C B is the midpoint of AC Prove: AE ≅ DC

Statement

Reasons

#71

Given: E is midpoint of AB AD ⊥ AB and BC ⊥ AB ∠1 ≅ ∠2 Prove: AD ≅ CB

Statement

Reasons

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