Objective Use CPCTC to prove parts of triangles are congruent.

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Presentation transcript:

Objective Use CPCTC to prove parts of triangles are congruent.

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.

You much prove SSS, SAS, ASA, AAS, or HL first, and THEN use CPCTC to show that corresponding parts of triangles are congruent. Remember!

Example 1: Engineering Application A and B are on the edges of a ravine. What is AB? 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.

Example 2: Proving Corresponding Parts Congruent Given: YW bisects XZ, XY  YZ. Prove: XYW  ZYW Z

Example 2 Continued WY ZW

Given: PR bisects QPS and QRS. Check It Out! Example 2 Prove: PQ  PS Given: PR bisects QPS and QRS.

Check It Out! Example 2 Continued PR bisects QPS and QRS QRP  SRP QPR  SPR Given Def. of  bisector RP  PR Reflex. Prop. of  ∆PQR  ∆PSR PQ  PS ASA CPCTC

Then look for triangles that contain these angles. Work backward when planning a proof. To show that ED || GF, look for a pair of angles that are congruent. Then look for triangles that contain these angles. Helpful Hint

Example 3: Using CPCTC in a Proof Prove: MN || OP Given: NO || MP, N  P

Example 3 Continued Statements Reasons 1. N  P; NO || MP 1. Given 2. NOM  PMO 2. Alt. Int. s Thm. 3. MO  MO 3. Reflex. Prop. of  4. ∆MNO  ∆OPM 4. AAS 5. NMO  POM 5. CPCTC 6. MN || OP 6. Conv. Of Alt. Int. s Thm.

2. Given: X is the midpoint of AC . 1  2 Lesson Quiz: Part II 2. Given: X is the midpoint of AC . 1  2 Prove: X is the midpoint of BD.

Lesson Quiz: Part II Continued 6. CPCTC 7. Def. of  7. DX = BX 5. ASA Steps 1, 4, 5 5. ∆AXD  ∆CXB 8. Def. of mdpt. 8. X is mdpt. of BD. 4. Vert. s Thm. 4. AXD  CXB 3. Def of  3. AX  CX 2. Def. of mdpt. 2. AX = CX 1. Given 1. X is mdpt. of AC. 1  2 Reasons Statements 6. DX  BX