Congruent Figures GEOMETRY LESSON 4 1 For help

Congruent Figures GEOMETRY LESSON 4 -1 (For help, go to page 24. ) Solve each equation. 1. x + 6 = 25 2. x + 7 + 13 = 33 3. 5 x = 540 4. x + 10 = 2 x 5. For the triangle at the right, use the Triangle Angle-Sum Theorem to find the value of y. 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 Solutions 1. Subtract 6 from both sides: x = 19 2. Combine like terms: x + 20 = 33; subtract 20 from both sides: x = 13 3. Divide both sides by 5: x = 108 4. Subtract x from both sides of x + 10 = 2 x: 10 = x, or x = 10 5. y + 40 + 90 = 180; combine like terms: y + 130 = 180; subtract 130 from both sides: y = 50 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 ABC QTJ. List the congruent corresponding parts. List the corresponding vertices in the same order. Angles: A Q B T C J List the corresponding sides in the same order. Sides: AB QT BC TJ AC 4 -1 QJ

Congruent Figures GEOMETRY LESSON 4 -1 XYZ KLM, m Y = 67, and m M = 48. Find m X. Use the Triangle Angle-Sum Theorem and the definition of congruent polygons to find m X + m Y + m Z = 180 Triangle Angle-Sum Theorem m Z = m M Corresponding angles of congruent triangles that are congruent m Z = 48 Substitute 48 for m M. m X + 67 + 48 = 180 m X + 115 = 180 m X = 65 Substitute. Simplify. Subtract 115 from each side. 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 Can you conclude that ABC CDE in the figure below? List corresponding vertices in the same order. If ABC CDE, then BAC DCE. The diagram above shows BAC DEC, not DCE. Corresponding angles are not necessarily congruent, therefore you cannot conclude that ABC CDE. 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 Show you can conclude that CNG DNG. List statements and reasons. Congruent triangles have three congruent corresponding sides and three congruent corresponding angles. Examine the diagram, and list the congruent corresponding parts for CNG and DNG. a. CG DG b. CN DN c. GN GN d. C D e. CNG DNG f. CGN DGN g. CNG DNG Given Reflexive Property of Congruence Given Right angles are congruent. If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent. (Theorem 4 -1. ) Definition of triangles 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 Pages 182 -185 Exercises 1. 2. CAB DAB; C D; ABC ABD; AC AD; AB AB CB DB GEF JHI; GFE JIH; EGF HJI; GE JH; EF HI FG IJ 5. ML 13. E, K, G, N 6. B 14. PO SI; OL ID; LY DE; PY SE 7. C 15. P L 8. J 9. KJB 16. 33 in. 10. CLM 17. 54 in. 3. BK 11. JBK 18. 105 4. CM 12. MCL 19. 77 4 -1 S; D; O Y I; E

Congruent Figures GEOMETRY LESSON 4 -1 20. 36 in. 27. Yes; all corr. sides and s are. 21. 34 in. 28. a. Given b. If || lines, then alt. int. 22. 75 23. 103 24. Yes; RTK UTK, R U (Given) RKT UKT If two s of a are to two s of another , the third s are. TR TU, RK UK (Given) TK TK (Reflexive Prop. of ) TRK TUK s ) (Def. of 25. No; the corr. sides are not are. Given If 2 s of one are to two s of another then 3 rd s are. Reflexive Prop. of Given s Def. of s . 26. No; corr. sides are not necessarily . 4 -1 c. d. e. f. g. ,

Congruent Figures GEOMETRY LESSON 4 -1 29. A and H; B and G; C and E; D and F 30. x = 15; t = 2 36. Answers may vary. Sample: It is important that PACH OLDE for the patch to completely fill the hole. 31. 5 32. m A=m D = 20 33. m B = m E = 21 37. Answers may vary. Sample: She could arrange them in a neat pile and pull out the ones of like sizes. 34. BC = EF = 8 38. JYB XCH 39. BCE ADE 40. TPK TRK 35. AC = DF = 19 4 -1 41. JLM NRZ; ZRN 42. Answers may vary. Sample: The die is a mold that is used to make items that are all the same size. 43. Answers may vary. Sample: TKR MJL: TK MJ; TR ML; KR JL; TKR MJL; TRK MLJ; KTR JML

Congruent Figures GEOMETRY LESSON 4 -1 44. a. Given b. If || lines, then alt. int. s are. c. If || lines, then alt. int. s are. d. Vertical s are . 48. a. 15 45. Answers may vary. Sample: Since the sum b. of the s of a is 180, and if 2 s of one are the same as 2 s of a second , then their sum subtracted from 180 has to be the same. e. Given 46. KL = 4; LM = 3; KM = 5 f. Given g. Def. of segment bisector h. Def. of 47. 2; either (3, 1) or (3, – 7) s 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 49. 1. 5 54. 50. 4. 25 51. 40 52. 72 55. 100 53. Answers may vary. Sample: 56. RS = PQ 57. 1 58. 12 59. AB GH 4 -1

Congruent Figures GEOMETRY LESSON 4 -1 In Exercises 1 and 2, quadrilateral WASH quadrilateral NOTE. 1. List the congruent corresponding parts. WA NO, AS OT, SH TE, WH NE; W N, A O, S T, H E 2. m O = m T = 90 and m H = 36. Find m N. 144 3. Write a statement of triangle congruence. Sample: DFH ZPR 4. Write a statement of triangle congruence. Sample: ABD CDB 5. Explain your reasoning in Exercise 4 above. Sample: Two pairs of corresponding sides and two pairs of corresponding angles are given. C A because all right angles are congruent. BD BD by the Reflexive Property of. ABD CDB by the definition of congruent triangles. 4 -1

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 (For help, go to Lesson 2 -5. ) What can you conclude from each diagram? 1. 2. 3. 4 -2

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 Solutions 1. According to the tick marks on the sides, AB tick marks on the angles, C F. DE. According to the 2. The two triangles share a side, so PR PR. According to the tick marks on the angles, QPR SRP and Q S. 3. According to the tick marks on the sides, TO NV. The tick marks on the angles show that M S. Since MO || VS, by the Alternate Interior Angles Theorem MON SVT. Since OV OV by the Reflexive Property, you can use the Segment Addition Property to show TV NO. 4 -2

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 Given: M is the midpoint of XY, AX Prove: AMX AMY AY Write a paragraph proof. Copy the diagram. Mark the congruent sides. You are given that M is the midpoint of XY, and AX AY. Midpoint M implies MX MY. AM AM by the Reflexive Property of Congruence, so AMX AMY by the SSS Postulate. 4 -2

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 AD BC. What other information do you need to prove ADC BCD? It is given that AD BC. Also, DC CD by the Reflexive Property of Congruence. You now have two pairs of corresponding congruent sides. Therefore: Solution 1: If you know AC by SSS. BD, you can prove Solution 2: If you know ADC by SAS. ADC BCD, you can prove 4 -2 BCD ADC BCD

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 Given: RSG RSH, SG SH From the information given, can you prove RSG RSH? Explain. Copy the diagram. Mark what is given on the diagram. It is given that RSG RSH and SG SH. RS RS by the Reflexive Property of Congruence. Two pairs of corresponding sides and their included angles are congruent, so RSG RSH by the SAS Postulate. 4 -2

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 Pages 189 -192 Exercises 6. Yes; AC DB (Given); 10. U, AE CE and BE DE (Def. of midpt. ); AEB 11. WU s CED (vert. are ) AEB CED by SAS. 12. X 1. SSS 2. cannot be proved 3. SAS 4. SSS 5. Yes; OB OB by Refl. Prop. ; BOP BOR since all rt. are ; OP OR (Given); the s are by SAS. s 7. a. Given b. Reflexive c. JKM d. LMK 8. WV, VU 9. 13. XZ, YZ 14. LG 15. T or RS 16. DC W 4 -2 V MN V WU CB

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 17. additional information not needed 18. Yes; ACB by SAS. 19. Yes; PVQ by SSS. EFD 22. ANG LOM KLJ MON; SSS 24. Not possible; need H P or DY TK. 25. JEF 26. BRT BRS; SSS 27. PQR NMO; SAS STR 20. No; need YVW ZVW or YW ZW. 21. Yes; NMO by SAS. 23. SVF or SFV; SSS 28. No; even though the s are , the sides may not be. RWT; SAS 4 -2 29. No; you would need H K or GI JL. 30. yes; SAS 31.

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 32. 34. Answers may vary. Sample: 35. (continued) b. 35. a–b. Answers may vary. Sample: 33. a. b. c. d. e. f. Vertical Given s are Def. of midpt. Given Def. of midpt. SAS . a. wallpaper designs; ironwork on a bridge; highway warning 36. signs produce a wellbalanced, symmetric appearance. In s construction, enhance designs. Highway warning signs are more easily identified if they are. s ISP by SAS. PSO; OSP 37. IP PO; ISP OSP by SSS. 4 -2

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 38. Yes; ADB CBD by SAS; ADB DBC because if || lines, then alt. int. s are. 41. 1. FG || KL (Given) 2. GFK FKL (If || lines, then alt. int. s are. ) 3. FG 39. Yes; ABC CDA by SAS; DAC ACB because if || lines, then alt. int. s are. KL (Given) 4. FK FK (Reflexive Prop. of ) 5. FGK (SAS) 40. No; ABCD could be a square with side 5 and EFGH could be a polygon with side 5 but no rt. s. 4 -2 KLF 42. AE and BD bisect each other, so AC CE and BC CD. ACB DCE because vert. s are ACB ECD by SAS. .

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 43. 45. D 51. FG 46. G 52. 47. C 48. [2] a. AB AB; Reflexive Prop. of 44. AM MB because M is the midpt. of AB. B AMC because all right s are. CM DB is given. AMC MBD by SAS. b. No; AB is not a corr. side. [1] one part correct 49. E 50. AB 4 -2 C 53. The product of the slopes of two lines is – 1 if and only if the lines are. 54. If x = 2, then 2 x = 4. If 2 x = 4, then x = 2. 55. If 2 x = 6, then x = 3. The statement and the converse are both true.

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 56. If x 2 = 9, then x = 3. The statement is true but the converse is false. 4 -2

Triangle Congruence by SSS and SAS GEOMETRY LESSON 4 -2 1. In VGB, which sides include B? BG and BV 2. In STN, which angle is included between NS and TN? N 3. Which triangles can you prove congruent? Tell whether you would use the SSS or SAS Postulate. APB XPY; SAS 4. What other information do you need to prove DWO DWG? If you know DO DG, the triangles are by SSS; if you know DWO DWG, they are by SAS. 5. Can you prove SED BUT from the information given? Explain. No; corresponding angles are not between corresponding sides. 4 -2

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 (For help, go to the Lesson 4 -2. ) In JHK, which side is included between the given pair of angles? 1. J and H 2. H and K In NLM, which angle is included between the given pair of sides? 3. LN and LM 4. NM and LN Give a reason to justify each statement. 5. PR PR 6. A 4 -3 D

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 Solutions 1. JH 2. HK 3. L 4. N 5. By the Reflexive Property of Congruence, a segment is congruent to itself. 6. By the Triangle Angle-Sum Theorem, the sum of the angles of any triangle is 180. If m C = m F = x and m B = m E = y, then m A = 180 – x – y = m D, so A = D. 4 -3

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 Suppose that F is congruent to C and I is not congruent to C. Name the triangles that are congruent by the ASA Postulate. The diagram shows N If F C, then F Therefore, FNI A C CAT D and FN G GDO by ASA. 4 -3 CA GD.

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 Write a paragraph proof. Given: A Prove: B, AP APX It is given that A APX BP BPY B and AP BP. BPY by the Vertical Angles Theorem. Because two pairs of corresponding angles and their included sides are congruent, APX BPY by ASA. 4 -3

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 Write a Plan for Proof that uses AAS. Given: B Prove: D, AB || CD ABC CDA Because AB || CD, BAC Interior Angles Theorem. DCA by the Alternate Then ABC CDA if a pair of corresponding sides are congruent. By the Reflexive Property, AC ABC CDA by AAS. 4 -3 AC so

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 Write a two-column proof that uses AAS. Given: B D, AB || CD Prove: ABC CDA Statements Reasons 1. B 1. Given D, AB || CD 2. BAC 3. AC 4. DCA CA ABC 2. If lines are ||, then alternate interior angles are. 3. Reflexive Property of Congruence CDA 4. AAS Theorem 4 -3

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 Pages 197 -201 Exercises 1. PQR VXW 2. ACB EFD N and 13. a. b. ASA O 5. yes 6. not possible UWV b. UW c. right 9. AAS 3. RS 4. 8. a. Reflexive d. Reflexive 10. ASA 14. B D 11. not possible 12. FDE DFE GHI; HGI 7. yes 15. MU UN 16. PQ QS 17. 4 -3 WZV WZY

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 18. a. Vert. s are . b. Given c. TQ QR d. AAS 19. PMO 20. UTS 21. ZVY 22. TUX 24. TXU ODE; ASA 25. The are not s are because the not included s. s NMO; ASA 26. Yes; if 2 s of a are to 2 s of another , RST; AAS then the 3 rd s are. So, an AAS proof can WVY; AAS be rewritten as an ASA proof. DEO; AAS 23. The s are not because no sides are. 4 -3 27. a. SRP b. PR c. alt. int. d. PR e. Reflexive 28. a. b. c. d. Given Def. of bis. Given Reflexive Prop. of e. AAS

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 29. a. Def. of b. All right c. QTP s are. STR d. Def. of midpt. e. AAS 30. 31. Yes; by AAS since MON QOP. 34. Yes; by ASA since BDH FDH by def. of bis. and DH DH by the Reflexive Prop. of. 32. Yes; by AAS since FGJ HJG because when lines are 35. Answers may vary. ||, then alt. int. s are Sample: and GJ GJ by the Reflexive Prop. of. 33. Yes; by ASA, since EAB DBC because || lines have corr. s. 4 -3 36. a. Check students’ work. b. Answers may vary; most likely ASA.

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 37. 38. AEB BEC ABC BAD AEB BEC ABD BAD ABD CBA BCD 39. They are ASA. CED, DEA, CDA, DCB, DCB, DAB, ADC 44. [2] a. RPQ SPQ, RQP SQP (Def. of bisector) 40. 13 20 41. b. ASA [1] one part correct 42. D 43. F bisectors; 4 -3

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 45. [4] a. Def. of midpt. b. Yes; JLM KGM because they are alt. int. s of || lines, and LMJ GMK because vertical s are. So the s are by ASA. c. Yes; if two s of one are , the third s are. [3] incorrect s to 2 s of another for part b or c, but otherwise correct [1] at least one part correct ONL 49. If corr. s are , then the lines are ||. 50. 56 photos [2] correct conclusions but incomplete explanations for parts b and c 46. 48. AC and CB MLN; SAS 47. not possible 4 -3 51. 36 photos 52. 60% more paper

Triangle Congruence by ASA and AAS GEOMETRY LESSON 4 -3 1. Which side is included between R and F in FTR? 2. Which angles in STU include US? S and U RF Tell whether you can prove the triangles congruent by ASA or AAS. If you can, state a triangle congruence and the postulate or theorem you used. If not, write not possible. 3. 4. GHI AAS PQR 5. not possible 4 -3 ABX AAS ACX

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 (For help, go to the Lesson 4 -1. ) In the diagram, JRC HVG. 1. List the congruent corresponding angles. 2. List the congruent corresponding sides. You are given that TIC LOK. 3. List the congruent corresponding angles. 4. List the congruent corresponding sides. 4 -4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 Solutions 1. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, J H, R V, and C G. 2. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, JR HV, RC VG, and JC HG. 3. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, T L, I O, and C K. 4. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, TI LO, IC OK, and TC LK. 4 -4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 What other congruence statements can you prove from the diagram, in which SL SR, and 1 2 are given? SC SC by the Reflexive Property of Congruence, and LSC RSC by SAS. 3 4 by corresponding parts of congruent triangles are congruent. When two triangles are congruent, you can form congruence statements about three pairs of corresponding angles and three pairs of corresponding sides. List the congruence statements. 4 -4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 (continued) Sides: SL SC CL Angles: 1 2 Given 3 4 Corresponding Parts of Congruent Triangles CLS SR SC CR CRS Given Reflexive Property of Congruence Other congruence statement In the proof, three congruence statements are used, and one congruence statement is proven. That leaves two congruence statements remaining that also can be proved: CLS CRS CL CR 4 -4

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 The Given states that DEG and DEF are right angles. What conditions must hold for that to be true? DEG and DEF are the angles the officer makes with the ground. So the officer must stand perpendicular to the ground, and the ground must be level. 4 -4

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

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 12. ATM RMT by SAS because ATM RMT by alt. int. s 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 15. PKL QKL by def. 18. The s are by SAS of bisect, and KL KL so the distance by Reflexive Prop. of , across the sinkhole is so the s are by SAS. 26. 5 yd by CPCTC. 16. KL KL by Reflexive Prop. of ; PL LQ by Def. of bis. ; KLP KLQ by Def. of ; the s are by SAS. 17. KLP KLQ because all rt s are ; KL KL by Reflexive Prop. of ; and PKL QKL by def. of bisect; the s are by ASA. 4 -4 19. a. b. c. d. e. Given Def. of All right s are. Given Def. of segment bis. f. Reflexive Prop. of g. SAS h. CPCTC

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. s are. 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 s and BD AC by def. of. 25. a. AP PB; AC BC b. The diagram is constructed in such a way that the s are by SSS. CPA CPB by CPCTC. Since these s are and suppl. , they are right s. Thus, CP is to. 23. e or b, d, c, f, a 4 -4

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

Using Congruent Triangles: CPCTC GEOMETRY LESSON 4 -4 1. What does “CPCTC” stand for? Corresponding parts of congruent triangles are congruent. Use the diagram for Exercises 2 and 3. 2. Tell how you would show ABM ACM. You are given two pairs of s and AM AM by the Reflexive Prop. , so ABM ACM by ASA. 3. Tell what other parts are congruent by CPCTC. AB AC, BM CM, B C Use the diagram for Exercises 4 and 5. 4. Tell how you would show RUQ TUS. You are given a pair of s and a pair of sides and RUQ TUS because vertical angles are , so RUQ TUS by AAS. 5. Tell what other parts are congruent by CPCTC. RQ TS, UQ US, R T 4 -4

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 (For help, go to the Lesson 3 -3. ) 1. Name the angle opposite AB. 2. Name the angle opposite BC. 3. Name the side opposite A. 4. Name the side opposite C. 5. Find the value of x. 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Solutions 1. The angle opposite AB is the angle whose side is not AB: C 2. The angle opposite BC is the angle whose side is not BC: A 3. The side opposite A is the side that is not part of A: BC 4. The side opposite C is the side that is not part of C: BA 5. By the Triangle Exterior Angle Theorem, x = 75 + 30 = 105°. 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Examine the diagram below. Suppose that you draw XB YZ. Can you use SAS to prove XYB XZB? Explain. It is given that XY XZ. By the definition of perpendicular, XBY = XBZ. By the Reflexive Property of Congruence, XB XB. However, because the congruent angles are not included between the congruent corresponding sides, the SAS Postulate does not apply. You cannot prove the triangles congruent using SAS. 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Explain why ABC is isosceles. ABC and XAB are alternate interior angles formed by XA, BC, and the transversal AB. Because XA || BC, ABC XAB. The diagram shows that XAB ACB. By the Transitive Property of Congruence, ABC ACB. You can use the Converse of the Isosceles Triangle Theorem to conclude that AB AC. By the definition of an isosceles triangle, isosceles. 4 -5 ABC is

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Suppose that m L = y. Find the values of x and y. MO LN The bisector of the vertex angle of an isosceles triangle is the perpendicular bisector of the base. x = 90 Definition of perpendicular m N = m L Isosceles Triangle Theorem m L = y Given m N = y Transitive Property of Equality m N + m NMO + m MON = 180 Triangle Angle-Sum Theorem y + 90 = 180 Substitute. 2 y + 90 = 180 Simplify. 2 y = 90 Subtract 90 from each side. y = 45 Divide each side by 2. Therefore, x = 90 and y = 45. 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Suppose the raised garden bed is a regular hexagon. Suppose that a segment is drawn between the endpoints of the angle marked x. Find the angle measures of the triangle that is formed. Because the garden is a regular hexagon, the sides have equal length, so the triangle is isosceles. By the Isosceles Triangle Theorem, the unknown angles are congruent. Example 4 found that the measure of the angle marked x is 120. The sum of the angle measures of a triangle is 180. If you label each unknown angle y, 120 + y = 180. 120 + 2 y = 180 2 y = 60 y = 30 So the angle measures in the triangle are 120, 30 and 30. 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Pages 213 -216 Exercises 1. a. RS 2. (continued) b. RS c. Given d. Def. of segment bisector d. Def. of bisector e. Reflexive Prop. of f. AAS e. Reflexive Prop. of f. SSS g. CPCTC 5. VY; VT = VX (Ex. 3) and UT = YX (Ex. 4), so VU = VY by the Subtr. Prop. of =. 6. Answers may vary. Sample: VUY; sopp. sides are. 7. x = 80; y = 40 2. a. KM b. KM c. By construction 3. VX; Conv. of the Isosc. Thm. 8. x = 40; y = 70 4. UW; Conv. of the Isosc. 9. x = 38; y = 4 Thm. 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 10. x = 4 1 ; y = 60 2 19. a. 22. 140 11. x = 36; y = 36 23. 6 12. x = 92; y = 7 24. x = 60; y = 30 13. 64 1 14. 2 2 30, 120 b. 5; 30, 60, 90, 120, 150 15. 42 16. 35 17. 150; 15 c. Check students’ work. 25. x = 64; y = 71 26. x = 30; y = 120 27. Two sides of a are if and only if the s opp. those sides are. 20. 70 28. 80, 20; 80, 50 18. 24, 48, 72, 96, 120 21. 50 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 29. a. isosc. s b. 900 ft; 1100 ft c. The tower is the bis. of the base of each. 30. No; the can be positioned in ways such that the base is not on the bottom. 31. 45; they are = and have sum 90. 32. Answers may vary. Sample: Corollary to Thm. 4 -3: Since XY YZ, X Z by Thm. 4 -3. YZ ZX, so Y X by Thm. 4 -3 also. By the Trans. Prop. , Y Z, so X Y Z. Corollary to Thm. 4 -4: Since X Z, XY YZ by Thm. 4 -4. Y X, so YZ ZX by Thm. 4 -4 also. By the Trans. Prop. XY ZX, so XY YZ ZX. 33. a. Given b. A c. Given d. ABE D DCE 34. m = 36; n = 27 35. m = 60; n = 30 36. m = 20; n = 45 37. (0, 0), (4, 4), (– 4, 0), (0, – 4), (8, 4), (4, 8) 38. (5, 0); (0, 5); (– 5, 5); (5, – 5); (0, 10); (10, 0) 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 39. (5, 3); (2, 6); (2, 9); (8, 3); (– 1, 6); (5, 0) 40. a. 25 b. 40; 100 c. Obtuse isosc. 2 of the s are and one is obtuse. ; 41. AC CB and ACD DCB are given. CD CD by the Refl. Prop. of , so ACD BCD by SAS. So AD DB by CPCTC, and CD bisects AB. Also ADC BDC by CPCTC, m ADC + m BDC = 180 by Add. Post. , so m ADC = m BDC = 90 by the Subst. Prop. So CD is the bis. of AB. 4 -5 42. The bis. of the base of an isosc. is the bis. of the vertex ; given isosc. ABC with bis. CD, ADC BDC and AD DB by def. of bis. Since CD CD by Refl. Prop. , ACD BCD by SAS. So ACD BCD by CPCTC, and CD bisects ACB.

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 43. a. 5 44. 0 < measure of base < 45 b. 45 < measure of base < 90 49. [2] a. 60; since m PAB = m PBA, and m PAB + m PBA = 120, m PAB = 60. b. 120; m APB = 60 so m PAB = 60. Since PAB and QAB are compl. , m QAB = 30. QAB is isosc. so m AQB = 120. 46. C 47. G 48. D [1] one part correct 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 50. RC = GV; RC GV by CPCTC since RTC GHV by ASA. 51. AAS 52. SSS 53. 24 sides 4 -5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4 -5 Use the diagram for Exercises 1– 3. 1. If m BAC = 38, find m C. 71 2. If m BAM = m CAM = 23, find m BMA. 90 3. If m B = 3 x and m BAC = 2 x – 20, find x. 25 4. Find the values of x and y. x = 60 y=9 5. ABCDEF is a regular hexagon. Find m BAC. 30 4 -5

Congruence in Right Triangles GEOMETRY LESSON 4 -6 (For help, go to the Lessons 4 -2 and 4 -3. ) Tell whether the abbreviation identifies a congruence statement. 1. SSS 2. SAS 3. SSA 4. ASA 5. AAS 6. AAA Can you conclude that the two triangles are congruent? Explain. 7. 8. 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 Solutions 1. SSS names the Side-Side Theorem. 2. SAS names the Side-Angle-Side Theorem. 3. SSA does not name a theorem or postulate. 4. ASA names the Angle-Side-Angle Postulate. 5. AAS names the Angle-Side Theorem. 6. AAA does not name a congruence theorem or postulate. 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 Solutions (continued) 7. The side that they share is congruent to itself by the Reflexive Property of Congruence. The two right angles are congruent to each other, and two other corresponding sides are marked congruent. The two triangles are congruent by SAS. 8. Two pairs of congruent sides are marked congruent. The included angles are congruent because vertical angles are congruent. Thus, the two triangles are congruent by SAS. 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 One student wrote “ CPA below. Is the student correct? Explain. MPA by SAS” for the diagram The diagram shows the following congruent parts. CA MA CPA PA MPA PA There are two pairs of congruent sides and one pair of congruent angles, but the congruent angles are not included between the corresponding congruent sides. The triangles are not congruent by the SAS Postulate, but they are congruent by the HL Theorem. 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 XYZ is isosceles. From vertex X, a perpendicular is drawn to YZ, intersecting YZ at point M. Explain why XMY XMZ. 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 Write a two–column proof. Given: ABC and DCB are right angles, AC Prove: ABC DCB Statements DB Reasons 1. ABC and DCB are right angles. 2. ABC and DCB are right triangles. 3. AC DB 4. BC CB 1. Given 5. If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. (HL Theorem). ABC DCB 2. Definition of a right triangle 3. Given 4. Reflexive Property of Congruence 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 Pages 219 -223 Exercises 1. ABC DEF by HL. 4. AEB DCB by HL. Both s are rt. s , Both s are rt. s. AC DF, and CB FE. AB BD and EB CB by the def. of midpt. 2. SPR QRP by HL. Both s are rt. s , SP 5. T and Q are rt. s. QR (given) and PR PR by the Reflexive 6. RX RT or XV TV Prop. of. 7. TY ER or RT YE 3. LMP OMN by HL. Both s are rt. s 8. Right s are needed, because vert. s are ; either A and G or LP NO, and LM AQC and GJC. OM. 4 -6 9. BC FA 10. RT NQ 11. a. Given b. Def. of rt. c. Reflexive Prop. of d. Given e. HL

Congruence in Right Triangles GEOMETRY LESSON 4 -6 12. a. suppl. s are rt. s b. Def. of rt. c. Given d. Reflexive Prop. of e. HL 13. PS PT so S T by the Isosc. Thm. PRS PRT by AAS. 15. Yes; PM PM and PMW is a rt. since JP || MW. b. IGH c. Def. of rt. d. I is the midpt. of HV. 16. a. Given b. Def. of c. MLJ and rt. s. d. Given e. LJ LJ f. HL 17. a. Given KJL are e. Def. of midpt. f. IGH ITV 18. HL; each rt. has a hyp. and side. 19. x = 3; y = 2 20. x = – 1; y = 3 14. Yes; RS TU and RT TV. 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 21. whether the 7 -yd side is the hyp. or a leg 22. ABX ABC or SAS BXC ADX; HL ADC; SSS 24. 23. a. Answers may vary. Sample: You could show that suppl. s AXB and AXD are. b. DXC; HL ABC ADC by SSS so BAC DAC by CPCTC. ABX ADX by SAS so AXB AXD. AXB is suppl. and to AXD so they are both rt. s. 4 -6 25. 26. 27.

Congruence in Right Triangles GEOMETRY LESSON 4 -6 28. 1. EB DB; A and 29. 1. LO bisects MLN, C are rt. s (Given) OM LM, ON LN, (Given) 2. BEA BDC are rt. s 2. M and N are rt. s (Def. of rt. ) (Def. of ) 3. B is the midpt. of 3. MLO NLO AC (Given) (Def. of bis. ) 4. AB BC 4. M N (Def. of midpt. ) (All rt. s are. ) 5. BEA BDC 5. LO LO (Reflexive (HL) Prop. of ) 6. LMO (AAS) 4 -6 LNO 30. Answers may vary. Sample: Measure 2 sides of the formed by the amp. and the platform’s corner. Since the s will be by HL or SAS, the s are the same.

Congruence in Right Triangles GEOMETRY LESSON 4 -6 31. a. 31. (continued) e. EGD HL. Both DE EG b. slope of DG = – 1; slope of GF = – 1; slope of GE = 1 c. EGD and EGF are rt. s. d. DE = 26 ; FE = 26 s EGF by are rt. s , FE, and EG. 32. An HA Thm. is the same as AAS with AAS corr. to the rt. , an acute , and the hyp. 33. Since BE EA and BE EC, AEB and CEB are both rt. s. AB BC because ABC is equilateral, and BE BE. AEB CEB by HL. 34. No; AB CB because AEB CEB, but doesn’t have to be to AB or to CB. 35. A 36. H 4 -6

Congruence in Right Triangles GEOMETRY LESSON 4 -6 37. D 38. [2] a. b. TFW TGW RFW and RGW are rt. s . 41. BC || AD because each slope = – 1. BT BA, BA AD because product of slopes is – 1. 42. If || lines then alt. int. are. s [1] one part correct 39. isosceles 43. If two lines are ||, then same-side int. s are suppl. 40. equilateral 44. Vert. s are . 45. If two lines are ||, then corr. s are. 4 -6 46. If two lines are ||, then corr. s are. 47. If two lines are ||, then same-side int. s are suppl.

Congruence in Right Triangles GEOMETRY LESSON 4 -6 For Exercises 1 and 2, tell whether the HL Theorem can be used to prove the triangles congruent. If so, explain. If not, write not possible. 1. 2. Not Yes; use the congruent possible hypotenuses and leg BC to prove ABC DCB For Exercises 3 and 4, what additional information do you need to prove the triangles congruent by the HL Theorem? 3. LMX LOX 4. AMD CNB LM LO AM CN or MD 4 -6 NB

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 (For help, go to the Lessons 1 -1 and 4 -3. ) 1. How many triangles will the next two figures in this pattern have? 2. Can you conclude that the triangles are congruent? Explain. a. AZK and DRS b. SDR and JTN c. ZKA and NJT 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 Solutions 1. For every new right triangle, segments connect the midpoint of the hypotenuse with the midpoints of the legs of the right triangle, creating two new triangles for every previous new triangle. The first figure has 1 triangle. The second has 1 + 2, or 3 triangles. The third has 3 + 4, or 7 triangles. The fourth will have 7 + 8, or 15 triangles. The fifth will have 15 + 16, or 31 triangles. 2. a. Two pairs of sides are congruent. The included angles are congruent. Thus, the two triangles are congruent by SAS. b. Two pairs of angles are congruent. One pair of sides is also congruent, and, since it is opposite a pair of corresponding congruent angles, the triangles are congruent by AAS. c. Since AZK Property of , DRS and SDR ZKA NJT. 4 -7 JTN, by the Transitive

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 Name the parts of their sides that DFG and Identify the overlapping triangles. Parts of sides DG and EG are shared by These parts are HG and FG, respectively. 4 -7 DFG and EHG share.

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 Write a Plan for Proof that does not use overlapping triangles. Given: ZXW YWX, ZWX Prove: ZW YX YXW Label point M where ZX intersects WY, as shown in the diagram. ZW YX by CPCTC if ZWM YXM. You can prove these triangles congruent using ASA as follows: Look at MWX. MW Theorem. MX by the Converse of the Isosceles Triangle Look again at ZWM and YXM. ZMW YMX because vertical angles are congruent, MW MX, and by subtraction ZWM YXM, so ZWM YXM by ASA. 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 Write a paragraph proof. Given: XW YZ, XWZ and YZW are right angles. Prove: XPW YPZ Plan: XPW YPZ by AAS if WXZ ZYW. These angles are congruent by CPCTC if XWZ YZW. These triangles are congruent by SAS. Proof: You are given XW YZ. Because XWZ and YZW are right angles, XWZ YZW. WZ ZW, by the Reflexive Property of Congruence. Therefore, XWZ YZW by SAS. WXZ ZYW by CPCTC, and XPW YPZ because vertical angles are congruent. Therefore, XPW YPZ by AAS. 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 Given: CA CE, BA DE Write a two-column proof to show that CBE Proof: Statements Plan: CBE CDA by CPCTC if CBE congruence holds by SAS if CB CD. Reasons 1. BCE DCA 2. CA CE, BA DE 3. CA = CE, BA = DE 4. CA – BA = CE – DE 5. CA – BA = CB, CE – DE = CD 6. CB = CD 7. CB CD 8. CBE CDA 9. CBE CDA. This 1. Reflexive Property of Congruence 2. Given 3. Congruent sides have equal measure. 4. Subtraction Property of Equality 5. Segment Addition Postulate 6. Substitution 7. Definition of congruence 8. SAS 9. CPCTC 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 Pages 226 -230 Exercises 1. M 6. 9. 7. 10. a. Given b. Reflexive Prop. of c. Given d. AAS e. CPCTC 2. DF 3. XY 4. 5. 8. 11. 4 -7 LQP PML; HL

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 16. AE DE if AEB 18. ADC EDG by DEC by AAS. AB ASA if A E. A DC and A D since and E are corr. 13. QDA UAD; SAS they are corr. parts of parts in ADB and ABC and DCB, EDF, which are 14. QPT RUS; AAS which are by HL. by SAS. 15. TD RO if TDI QEU by 19– 22. Answers may ROE by AAS. TID 17. QET SAS if QT QU. QT vary. Samples REO if TEI and QU are corr. parts are given. RIE. TEI RIE of QTB and QUB 19. by SSS. which are by ASA. 12. RST UTS; SSS 20. 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 21. a. 23. ACE BCD by ASA; AC BC, A B (Given) C C (Reflexive Prop. of ) ACE BCD (ASA) 25. m m m 24. WYX ZXY by HL; WY YX, ZX YX, WX ZY (Given) WYX and ZXY are rt. s (Def. of ) XY XY (Reflexive Prop. of ) WYX ZXY (HL) 26. b. 22. a. b. 1 = 56; m 3 = 34; m 5 = 22; m 7 = 34; m 9 = 112 ABC FCG; ASA 27. a. Given b. Reflexive Prop. of c. Given d. 4 -7 2 = 56; 4 = 90; 6 = 34; 8 = 68; ETI

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 27. (continued) e. IRE 28. a. Given 28. (continued) i. DEC b. Def. of f. CPCTC j. Vert. s are c. Def. of rt. g. Given k. AAS d. Given h. All rt. s are . l. AE e. BC i. BC TDI m. CPCTC f. Reflexive j. DE ROE g. HL k. CPCTC h. CPCTC 4 -7 .

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 29– 30. Proofs may vary. Samples are given. 29. It is given that 1 2 and 3 4. Since QB QB by the Reflexive Prop. of , QTB QUB by ASA. So QT QU by CPCTC. Since QE QE by the Reflexive Prop. of , then QET QEU by SAS. 30. 1. AD ED (Given) 2. D is the midpt. of BF. (Given) 30. (continued) 7. GDE CDA (Vert. s are. ) 8. 3. FD DB (Def. of midpt. ) 4. FDE ADB (Vert. s are. ) 5. FDE (SAS) BDA 6. E A (CPCTC) 4 -7 ADC (ASA) 31. a. AD AB AE DE BC; DC; EB EDG

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 32. 1. AC EC; 31. (continued) CB CD (Given) b. Use DB DB (refl. ) and alt. int. s to 2. C C (Reflexive show ADB Prop. of ) CBD (ASA). AB DC and AD BC 3. ACD ECB (CPCTC). AEB (SAS) CED (ASA) and AED CEB 4. A E (CPCTC) (ASA). Then AE EC and DE EB (CPCTC). 33. PQ RQ and PQT RQT by Def. of bisector. QT QT so PQT RQT by SAS. P R by CPCTC. QT bisects VQS so VQT SQT and PQT and RQT are both rt. s. So VQP SQR since they are s. compl. of PQV RQS by ASA so QV QS by CPCTC. 34. C 35. F 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 36. A 38. [4] a. HL 37. [2] a. HBC HED b. HB HE by CPCTC if HBC HED by ASA. Since BDC CED by AAS, then DBC CED by CPCTC and CHB DHE because s vertical are. [1] one part correct b. c. x = 30. In ADC m A + m ADC + m ACD = 180. Substituting, 90 + x + x = 180. Solving, x = 30. d. 120; it is suppl. to a 60° e. 6 m; DC = 2(AD) 4 -7

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 38. (continued) [3] 4 parts answered correctly 45. y – 0 = – 40. 46– 48. Eqs. may vary, depending on pt. chosen. [2] 3 parts answered correctly [1] 2 parts answered correctly 39. a. right b. 1 (x – 0) 3 46. y – 4 = 2(x – 1) 41. 47. y + 5 = 1 2 42. y + 6 = (x – 2) c. Reflexive 43. y – 5 = 1(x – 0) d. HL 44. y – 6 = – 2(x + 3) 4 -7 5 (x – 3) 3 48. y + 3 = – 5 (x + 4) 6

Using Corresponding Parts of Congruent Triangles GEOMETRY LESSON 4 -7 XY 1. Identify any common sides and angles in AXY and BYX. For Exercises 2 and 3, name a pair of congruent overlapping triangles. State theorem or postulate that proves them congruent. 2. 3. KSR SAS 4. Plan a proof. Given: AC BD, AD Prove: XD XC BC MRS GHI ASA IJG XD XC by CPCTC if DXA CXB. This congruence holds by AAS if BAD ABC. Show BAD ABC by SSS. 4 -7

Congruent Triangles GEOMETRY CHAPTER 4 Page 236 1. PAY APL 2. ONE OSE 3. SAS 9. Answers may vary. 11. No; the lengths may Sample: The corr. sides be different. of the two s may not be. 12. 36 10. Answers may vary. Sample: 4. HL 5. not possible 6. SSS 7. ASA 8. AAS CE HD; CO HF; EO DF; C H; E D; O F 4 -A 13. AT || GS, so ATG SGT because they are alt. int. s. It is given that AT GS, and GT GT by the Reflexive Prop. of , so GAT TSG by SAS.

Congruent Triangles GEOMETRY CHAPTER 4 17. Answers may vary. 14. Since LN bisects Sample: OLM and ONM, OLN MLN and ONL MNL. LN LN by the Reflexive Prop. of , so OLN MLN by ASA. ABC CDA 15. CFE DEF; SSS 16. TQS TRA; SAS 4 -A