Using Congruent Triangles: CPCTC GEOMETRY LESSON 4-4 Pages 204-208 Exercises 1. PSQ SPR; SQ RP; PQ SR 2. AAS; ABC EBD; A E; CB DB; DE CA by CPCTC 3. SAS; KLJ OMN; K O; J N; KJ ON by CPCTC 4. SSS; HUG BUG; H B; HUG BUG; UGH UGB by CPCTC 5. They are ; the are by AAS, so all corr. ext. are also . 6. a. SSS b. CPCTC 7. ABD CBD by ASA because BD BD by Reflexive Prop. of ; AB CB by CPCTC. 8. MOE REO by SSS because OE OE by Reflexive Prop. of ; M R by CPCTC. s 9. SPT OPT by SAS because TP TP by Reflexive Prop. of ; S O by CPCTC. 10. PNK MNL by SAS because KNP LNM by vert. are ; KP LM by CPCTC. 11. CYT RYP by AAS; CT RP by CPCTC. s 4-4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4-4 12. ATM RMT by SAS because ATM RMT by alt. int. are ; AMT RTM by CPCTC. 13. Yes; ABD CBD by SSS so A C by CPCTC. 14. a. Given b. Given c. Reflexive Prop. of d. AAS s 15. PKL QKL by def. of bisect, and KL KL by Reflexive Prop. of , so the are by SAS. 16. KL KL by Reflexive Prop. of ; PL LQ by Def. of bis.; KLP KLQ by Def. of ; the are by SAS. 17. KLP KLQ because all rt are ; KL KL by Reflexive Prop. of ; and PKL QKL by def. of bisect; the are by ASA. 18. The are by SAS so the distance across the sinkhole is 26.5 yd by CPCTC. 19. a. Given b. Def. of c. All right are . d. Given e. Def. of segment bis. f. Reflexive Prop. of g. SAS h. CPCTC s s s s s 4-4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4-4 20. ABX ACX by SSS, so BAX CAX by CPCTC. Thus AX bisects BAC by the def. of bisector. 21. Prove ABE CDF by SAS since AE FC by subtr. 22. Prove KJM QPM by ASA since P J and K Q by alt. int. are . 23. e or b, e or b, d, c, f, a s 24. BA BC is given; BD BD by the Reflexive Prop. of and since BD bisects ABC, ABD CBD by def. of an bisector; thus, ABD CBD by SAS; AD DC by CPCTC so BD bisects AC by def. of a bis.; ADB CDB by CPCTC and ADB and CDB are suppl.; thus, ADB and CDB are right and BD AC by def. of . 25. a. AP PB; AC BC b. The diagram is constructed in such a way that the are by SSS. CPA CPB by CPCTC. Since these are and suppl., they are right . Thus, CP is to . s 4-4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4-4 26. 1. PR || MG; MP || GR (Given) 2. Draw PG. (2 pts. determine a line.) 3. RPG PGM and RGP GPM (If || lines, then alt. int. are .) 4. PGM GPR (ASA) A similar proof can be written if diagonal RM is drawn. 27. Since PGM GPR (or PMR GRM), then PR MG and MP GR by CPCTC. 28. C 29. C s 30. D 31. B 32. C 33. [2] a. KBV KBT; yes; SAS b. CPCTC [1] one part correct 34. ASA 35. AAS 4-4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4-4 36. 95; 85 37. The slope of line m is the same as the slope of line n. 38. not possible 39. not possible 4-4