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Chapter 4

Chapter 4 . Section 7 : Triangle Congruence: CPCTC. Objectives. Use CPCTC to prove parts of triangles are congruent. . What is CPCTC?.

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Chapter 4

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  1. Chapter 4 Section 7 : Triangle Congruence: CPCTC

  2. Objectives Use CPCTC to prove parts of triangles are congruent.

  3. What is CPCTC? 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.

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

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

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

  7. Z Example 3Proofs Given: YW bisects XZ, XY YZ. Prove: XYW  ZYW

  8. solution

  9. Example 4 Given: PR bisects QPS and QRS. Prove: PQ  PS

  10. QRP SRP QPR  SPR PR bisects QPS and QRS RP PR Reflex. Prop. of  Def. of  bisector Given ∆PQR  ∆PSR ASA PQPS CPCTC solutions

  11. Student guided practice Do problems 2 and3 in your book page 270

  12. Given:NO || MP, N P Prove:MN || OP Example 5

  13. 1. N  P; NO || MP 3.MO  MO 6.MN || OP solution 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.

  14. Given:J is the midpoint of KM and NL. Prove:KL || MN Example 6

  15. 1.J is the midpoint of KM and NL. 2.KJ  MJ, NJ  LJ 6.KL || MN solution Statements Reasons 1. Given 2. Def. of mdpt. 3. KJL  MJN 3. Vert. s Thm. 4. ∆KJL  ∆MJN 4. SAS Steps 2, 3 5. LKJ  NMJ 5. CPCTC 6. Conv. Of Alt. Int. s Thm.

  16. Student guided practice Do problem 4 in your book page 271

  17. Example 7 Given: D(–5, –5), E(–3, –1), F(–2, –3), G(–2, 1), H(0, 5), and I(1, 3) Prove: DEF  GHI

  18. solution Step 1 Plot the points on a coordinate plane.

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

  20. solution So DEGH, EFHI, and DFGI. Therefore ∆DEF  ∆GHI by SSS, and DEF  GHI by CPCTC.

  21. Example 8 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

  22. solution 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.

  23. Student guided practice Do problems 7-13 in your book page 271

  24. Closure 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

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