Chapter 4 Section 7 : Triangle Congruence: CPCTC

Use CPCTC to prove parts of triangles are congruent.

Objectives

CPCTC is an abbreviation for the phrase “Corresponding Parts of Congruent Triangles are Congruent.” It can be used as a justification in a proof after you have proven two triangles congruent.

What is CPCTC?

SSS, SAS, ASA, AAS, and HL use corresponding parts to prove triangles congruent. CPCTC uses congruent triangles to prove corresponding parts congruent.

Remember !!!

A and B are on the edges of a ravine. What is AB?

Example 1: Engineering Application

One angle pair is congruent, because they are vertical angles. Two pairs of sides are congruent, because their lengths are equal.

Therefore the two triangles are congruent by SAS. By CPCTC, the third side pair is congruent, so AB = 18 mi.

A landscape architect sets up the triangles shown in the figure to find the distance JK across a pond. What is JK?

Example 2

One angle pair is congruent, because they are vertical angles.

Two pairs of sides are congruent, because their lengths are equal. Therefore the two triangles are congruent by SAS. By CPCTC, the third side pair is congruent, so JK = 41 ft.

Given: YW bisects XZ, XY YZ. Prove: XYW ZYW

Example 3Proofs

Z

solution

Given: PR bisects QPS and QRS. Prove: PQ PS

Example 4

solutionsPR bisects QPS and QRS

QRP SRPQPR SPR

Given Def. of bisector

RP PR

Reflex. Prop. of

∆PQR ∆PSR

PQ PS

ASA

CPCTC

Do problems 2 and3 in your book page 270

Student guided practice

Example 5

Prove: MN || OP Given: NO || MP, N P

solution

5. CPCTC5. NMO POM

6. Conv. Of Alt. Int. s Thm.

4. AAS4. ∆MNO ∆OPM

3. Reflex. Prop. of

2. Alt. Int. s Thm.2. NOM PMO

1. Given

ReasonsStatements

3. MO MO

6. MN || OP

1. N P; NO || MP

Example 6

Prove: KL || MN Given: J is the midpoint of KM and NL.

solution

5. CPCTC5. LKJ NMJ6. Conv. Of Alt. Int. s Thm.

4. SAS Steps 2, 34. ∆KJL ∆MJN

3. Vert. s Thm.3. KJL MJN

2. Def. of mdpt.

1. GivenReasonsStatements

6. KL || MN

1. J is the midpoint of KM and NL. 2. KJ MJ, NJ LJ

Do problem 4 in your book page 271

Student guided practice

Given: D(–5, –5), E(–3, –1), F(–2, –3), G(–2, 1), H(0, 5), and I(1, 3)

Prove: DEF GHI

Example 7

Step 1 Plot the points on a coordinate plane.

solution

Step 2 Use the Distance Formula to find the lengths of the sides of each triangle.

solution

So DE GH, EF HI, and DF GI. Therefore ∆DEF ∆GHI by SSS, and DEF

GHI by CPCTC.

solution

Given: J(–1, –2), K(2, –1), L(–2, 0), R(2, 3), S(5, 2), T(1, 1)

Prove: JKL RST Solution: Step 1 Plot the points on a coordinate plane

Example 8

Step 2 Use the Distance Formula to find the lengths of the sides of each triangle.

RT = JL = √5, RS = JK = √10, and ST = KL = √17.

So ∆JKL ∆RST by SSS. JKL RST by CPCTC.

solution

Do problems 7-13 in your book page 271

Student guided practice

Today we saw CPTCP and how we can prove corresponding parts to corresponding triangles

Next class we are going to learn about Introduction to coordinate proofs

Closure