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Use CPCTC to prove parts of triangles are congruent.

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. Remember!.

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Use CPCTC to prove parts of triangles are congruent.

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  1. Objective Use CPCTC to prove parts of triangles are congruent.

  2. CPCTCis 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.

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

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

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

  6. ZW WY Example 2 Continued

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

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

  9. Helpful Hint 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.

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

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

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

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

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