Proving Triangles Congruent

SSSSSSIf we can show all 3 pairs of corr.If we can show all 3 pairs of corr.

sides are congruent, the triangles sides are congruent, the triangles have to be congruent.have to be congruent.

SASSASShow 2 pairs of sides and the Show 2 pairs of sides and the

included angles are congruent and included angles are congruent and the triangles have to be congruent.the triangles have to be congruent.

IncludedIncludedangleangle

Non-includedNon-includedanglesangles

This is called a common side.This is called a common side.It is a side for both triangles.It is a side for both triangles.

We’ll use the reflexive property.We’ll use the reflexive property.

Common sideCommon side

SSSSSS

Parallel linesParallel linesalt int anglesalt int angles

Common sideCommon side

SASSAS

Vertical anglesVertical angles

SASSAS

ASA, AAS and HL ASA, AAS and HL

ASA – 2 anglesASA – 2 anglesand the included sideand the included side

AA

SS

AAAAS – 2 angles andAAS – 2 angles andThe non-included sideThe non-included side AAAA

SS

HLHL ( hypotenuse leg ) is used ( hypotenuse leg ) is usedonly with right triangles, BUT, only with right triangles, BUT,

not all right triangles. not all right triangles.

HLHL ASAASA

SOME REASONS WE’LL BE USINGSOME REASONS WE’LL BE USING

• DEF OF MIDPOINTDEF OF MIDPOINT

• DEF OF A BISECTORDEF OF A BISECTOR

• VERT ANGLES ARE CONGRUENTVERT ANGLES ARE CONGRUENT

• DEF OF PERPENDICULAR BISECTORDEF OF PERPENDICULAR BISECTOR

• REFLEXIVE PROPERTY (COMMON SIDE)REFLEXIVE PROPERTY (COMMON SIDE)

• PARALLEL LINES ….. ALT INT ANGLESPARALLEL LINES ….. ALT INT ANGLES

Proof

• 1) O is the midpoint of MQ and NP

• 2)

• 3)

• 4)

• 1) Given

• 2) Def of midpoint

• 3) Vertical Angles

• 4) SAS

Given: O is the midpoint of MQ and NP

Prove: MON POQ

MON POQ

,MO OQ NO OP MON POQ

AA BB

CC

11 22

Given: CX bisects ACBGiven: CX bisects ACB A A ˜ B˜ BProve: Prove: ∆ACX∆ACX ˜ ∆BCX˜ ∆BCX

XX

====

AASAAS

PPAAAASS∆’∆’ss

CX bisects ACB GivenCX bisects ACB Given 1 = 2 1 = 2 Def Def of angle biscof angle bisc A = B GivenA = B Given CX = CX Reflexive PropCX = CX Reflexive Prop∆∆ACX ˜ ∆BCX AASACX ˜ ∆BCX AAS==

Proof

• 1)

• 2)

• 3)

• 1) Given

• 2) Reflexive Property

• 3) SSS

Given:

Prove: ABC ADC

ABC ADC AC AC

,AB CD BC AD

,AB CD BC AD

Proof

• 1)

• 2)

• 3)

• 4)

• 1) Given

• 2) Alt. Int. <‘s Thm

• 3) Reflexive Property

• 4) SAS

ABD CDB

ABD CDB

, ||AD CB AD CB

, ||AD CB AD CB

DB DBADB CBD

Given:

Prove: