Aim: Triangle Congruence - ASA Course: Applied Geometry

Do Now:

Aim: How to prove triangles are congruent using a 3rd shortcut: ASA.

Given:T is the midpoint of PQ, PQ bisects RS, and RQ SP. Explain how RTQ STP.

T

P S

R Q

Aim: Triangle Congruence - ASA Course: Applied Geometry

Do Now

You are given:T is the midpoint of PQ, PQ bisects RS, and RQ SP. Explain how RTQ STP.

RQ SP – we’re told so

PT TQ – a midpoint of a segment cuts the segment into two congruent parts

RT TS – a bisector divides a segment into 2 congruent parts

RTQ STP because of SSS SSS

(S S)

(S S)

(S S)

T

P S

R Q

Aim: Triangle Congruence - ASA Course: Applied Geometry

Sketch 14 – Shortcut #3

ASA ASAASA ASA

Copied 2 angles and included side:

BC B’C’, B B’, C C’

Copied 2 angles and included side:

BC B’C’, B B’, C C’

Shortcut for proving congruence in triangles:

Measurements showed:

B

A

C

B’

A‘

C’

ABC A’B’C’ABC A’B’C’

Aim: Triangle Congruence - ASA Course: Applied Geometry

Angle-Side-Angle

III. ASA = ASATwo triangles are congruent if two angles and the included side of one triangle are equal in measure to two angles and the

included side of the other triangle.A

B B’C C’

A’

If A = A', AB = A'B', B = B', then ABC = A'B'C'

If ASA ASA , then the triangles are congruent

Aim: Triangle Congruence - ASA Course: Applied Geometry

Model Problems

Is the given information sufficient to prove congruent triangles?

F

D EA B

C

DA B

C

E

C

D

B

A

YES

YES

NO

Aim: Triangle Congruence - ASA Course: Applied Geometry

Model Problems

Name the pair of corresponding sides that would have to be proved congruent in order to prove that the triangles are congruent by ASA.

D C

A B

F

D

C

B

A

D C

A B

DCA CAB

DFA BFC

DB DB

Aim: Triangle Congruence - ASA Course: Applied Geometry

Model Problem

CD and AB are straight lines which intersect at E. BA bisects CD. AC CD, BD CD.Explain how ACE BDE using ASA

CE ED – bisector cuts segment into 2 parts

C D – lines form right angles and all right angles are & equal 90o

ACE BDE because of ASA ASA

(S S)

(A A)

E D

B

C

A

CEA BED – intersecting straight lines form vertical angles which are opposite and (A A)

Aim: Triangle Congruence - ASA Course: Applied Geometry

Model Problem

1 2, D is midpoint of EC, 3 4.Explain how AED BCD using ASA

ED DC – a midpoint of a segment cuts the segment into two congruent parts

1 2 – Given: we’re told so

AED BCD because of ASA ASA

(S S)

(A A)

3 4 – Given: we’re told so (A A)

2431

A

DE C

B

Aim: Triangle Congruence - ASA Course: Applied Geometry

Model Problem

DA is a straight line, E B, ED AB, FD DE, CA ABExplain how DEF ABC using ASA

EDF BAC - lines form right angles and all right angles are & equal 90o

DEF ABC because of ASA ASA

(A A)

D

BA

F

E

C

E B – Given: we’re told so (A A)

ED AB – Given: we’re told so (S S)

Aim: Triangle Congruence - ASA Course: Applied Geometry