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# Objectives: - Identify & use the SSS, SAS, and ASA Congruence Postulates and the AAS and HL...

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Objectives: Identify & use the SSS, SAS, and ASA Congruence Postulates and the AAS and HL Congruence Theorems. Use counterexamples to prove that other side and angle combinations cannot be used to prove triangle congruence. 4.3 Analyzing Triangle Congruence Warm-Up: Which pair of triangles could use the ASA to prove congruency?
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Objectives:- Identify & use the SSS, SAS, and ASA

Congruence Postulates and the AAS and HL Congruence Theorems.

- Use counterexamples to prove that other side and angle combinations cannot be used to prove triangle congruence.

4.3 Analyzing Triangle Congruence

Warm-Up:

Which pair of triangles could use the ASA to prove congruency?

AAA combination – three angles

AAS combination – two angles and a side that is not between them

SSA combination- two sides and an angle that is not between them (that is, an angle opposite one of the two sides).

Valid or Not?:

AAA combination – three anglesValid or Not?:you can rule out the AAA combination by finding a counterexample for the following conjecture:

Conjecture: If the three angles of one triangle are congruent to the three angles of another triangle, then the triangles are congruent.

Counterexample: In the triangles below there are three pairs of congruent angles, but the two triangles are not congruent. Therefore the conjecture is false.

SSA combination- two sides & an angle that is not between them

Valid or Not?:you can rule out the SSA combination by finding a counterexample for the following conjecture:

Conjecture: If two sides and an angle that is not in between them of one triangle are congruent to two sides and an angle that is not in between them another triangle, then the triangles are congruent.Counterexample: In the triangles below the sides and angles are congruent, but the triangles are not congruent.

AAS combination – two angles & a side that is not between them

Valid or Not?:

The AAS combination can be converted to the ASA combination:The triangles below are an AAS combination. To convert this to an ASA combination, find the measures of the third angle in each triangle.A

B C E

D

F

𝟕𝟓𝟎𝟕𝟓𝟎

𝟔𝟎𝟎

𝟔𝟎𝟎

A

B C E

D

F𝟒𝟓𝟎𝟒𝟓𝟎𝟔𝟎𝟎

𝟔𝟎𝟎

Valid or Not?:

The two triangles below represent a version of the AAS, but the triangles are not congruent. There is an important difference between the two triangles. What is it?A

B C E

D

F𝟒𝟓𝟎𝟒𝟓𝟎

𝟔𝟎𝟎𝟔𝟎𝟎

88

If two angles and a non-included side of one triangle are congruent to the corresponding angles and non-included side of another triangle, then the triangles are congruent.

AAS (Angle-Angle-Side) Congruence Theorem:

For an AAS combination to be used, the congruent parts must correspond. Notice carefully the wording of the following theorem:

Example: Which pairs of congruent triangles can be proven to be congruent by the AAS Congruence Theorem?

If the hypotenuse and a leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, then the two triangles are congruent.

HL (Hypotenuse-Leg) Congruence Theorem

Examples: Determine whether the given combination of angles and sides determined a unique triangle. If so, identify the theorem or postulate that supports your answer.

∆ABC; AB=6, m<B= and m<A=

∆DEF; DE=5, EF=7 and m<F=

∆MNO; MN=8, MO=10, and m<N=

∆JKL; m<J= m<K=and m<L=

∆PQR; PQ=12, m<P= and m<R=

Textbook pages 230-232 Numbers 6-8, 10 -19

Homework:

4.3 Practice Worksheet

Objectives:- Use definitions & postulates to

prove a pair of triangles are congruent.

4.3 Triangle Congruence Proofs

Warm-Up:

Are the lines Parallel or curved?

Given:

Match the Corresponding Parts AB DE

Therefore ∆BAC ∆EDF

B

A

D

CAC DF<A <D

E

F

<A

<B

<C

AB

BC

AC

<D

DE

EF

<E

<F

DF

YW bisects <XYZ

Given: Prove:

XY YZ

∆YWX ∆YWZ

STATEMENTS REASONS

1. 1.

2. <1 <2

3. YW YW

4. ∆YWX ∆YWZ

2.

3.

4.

Y

X ZW

1 2

3 45 6

AB EDGiven: Prove:

CE CB

∆ABC ∆DEC

STATEMENTS REASONS

1. 1.

2.

3.

4. ∆ABC ∆DEC

2. If lines are || then alt int <‘s are

3. Vertical <‘s are

4.

A B

C

E D

4 6

5

1 3

2

HJ bisects IKGiven: Prove:

HJ IK

∆HJI ∆HJK

STATEMENTS REASONS

1. 1.

2.

3.

4. J is the midpoint of IK

2. Def of ⊥

3. All right <‘s are

4.

H

I KJ

6 5

1 2 3 4

5. IJ JK

6. HJ HJ

7. ∆HJI ∆HJK

5.

6.

7.

AB DCGiven: Prove:

AB DC

STATEMENTS REASONS

1. 1.

2. <3 <6

3.

2. If lines are || then alt int <‘s are

3. ref.

4.

A B

CD

4

65

1

32

ME = TAGiven: Prove:

<M <T

∆MEN ∆TAN

STATEMENTS REASONS

1. 1.

2.

3.

4.

2.

3.

4.

A

S N

M

T

E

QU & RT bisect eachother

Given: Prove:

∆QRS ∆TUS

STATEMENTS REASONS

1. 1.

2.

3.

4.

2.

3.

4.

Q T

S

R U

4

6

51

3

2

Homework:Practice Worksheet

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