1. Angle Relationships in Triangles Holt Geometry Lesson Presentation Lesson Presentation Holt McDougal Geometry

2. Review Topics Vertical Angles Parallel Lines cut by a Transversal Linear Angles (Straight Angle) Complementary and Supplementary Angles Reflexive, Symmetric and Transitive Properties Congruence Definition of Bisector and Midpoint

3. An auxiliary line is a line that is added to a figure to aid in a proof. An auxiliary line used in the Triangle Sum Theorem

4. A corollary is a theorem whose proof follows directly from another theorem. Here are two corollaries to the Triangle Sum Theorem.

5. The interior is the set of all points inside the figure. The exterior is the set of all points outside the figure. Interior Exterior

6. An interior angle is formed by two sides of a triangle. An exterior angle is formed by one side of the triangle and extension of an adjacent side. Interior Exterior 4 is an exterior angle. 3 is an interior angle.

7. Each exterior angle has two remote interior angles. A remote interior angle is an interior angle that is not adjacent to the exterior angle. Interior Exterior 3 is an interior angle. 4 is an exterior angle. The remote interior angles of 4 are 1 and 2.

9. Find mB. Example 3: Applying the Exterior Angle Theorem mA + mB = mBCD Ext.  Thm. 15 + 2x + 3 = 5x – 60 Substitute 15 for mA, 2x + 3 for mB, and 5x – 60 for mBCD. 2x + 18 = 5x – 60 Simplify. 78 = 3x Subtract 2x and add 60 to both sides. 26 = x Divide by 3. mB = 2x + 3 = 2(26) + 3 = 55°

10. Find mACD. Example 3 mACD = mA + mB Ext.  Thm. 6z – 9 = 2z + 1 + 90 Substitute 6z – 9 for mACD, 2z + 1 for mA, and 90 for mB. 6z – 9 = 2z + 91 Simplify. 4z = 100 Subtract 2z and add 9 to both sides. z = 25 Divide by 4. mACD = 6z – 9 = 6(25) – 9 = 141°

12. Example 4 Find mP and mT. P  TP  T mP = mT 2x 2 = 4x 2 – 32 –2x 2 = –32 x 2 = 16 So mP = 2x 2 = 2(16) = 32°. Since mP = mT, mT = 32°. Third s Thm. Def. of  s. Substitute 2x 2 for mP and 4x 2 – 32 for mT. Subtract 4x 2 from both sides. Divide both sides by -2.

13. Practice 1. Find mABD. 2. Find mN and mP. 124° 75°; 75°

14. Congruent Triangles Holt Geometry Lesson Presentation Lesson Presentation Holt McDougal Geometry

15. Geometric figures are congruent if they are the same size and shape. Corresponding angles and corresponding sides are in the same position in polygons with an equal number of sides. Two polygons are congruent polygons if and only if their corresponding sides are congruent. Thus triangles that are the same size and shape are congruent.

17. When you write a statement such as ABC  DEF, you are also stating which parts are congruent. Helpful Hint

18. EXAMPLE Given: ∆PQR  ∆STW Identify all pairs of corresponding congruent parts. Angles:  P   S,  Q   T,  R   W Sides: PQ  ST, QR  TW, PR  SW

19. If polygon LMNP  polygon EFGH, identify all pairs of corresponding congruent parts. Example 1 Angles:  L   E,  M   F,  N   G,  P   H Sides: LM  EF, MN  FG, NP  GH, LP  EH

20. Example 2 Given: ∆ABC  ∆DBC. Find the value of x. BCA and BCD are rt. s. BCA  BCD mBCA = mBCD (2x – 16)° = 90° 2x = 106 x = 53 Def. of  lines. Rt.   Thm. Def. of  s Substitute values for mBCA and mBCD. Add 16 to both sides. Divide both sides by 2.

21. Example 2B Given: ∆ABC  ∆DBC. Find mDBC. mABC + mBCA + mA = 180° mABC + 90 + 49.3 = 180 mABC + 139.3 = 180 mABC = 40.7 DBC  ABC mDBC = mABC ∆ Sum Thm. Substitute values for mBCA and mA. Simplify. Subtract 139.3 from both sides. Corr. s of  ∆s are . Def. of  s. mDBC  40.7° Trans. Prop. of =

22. Given: ∆ABC  ∆DEF Example 3 Find the value of x. 2x – 2 = 6 2x = 8 x = 4 Corr. sides of  ∆s are . Add 2 to both sides. Divide both sides by 2. AB  DE Substitute values for AB and DE. AB = DE Def. of  parts.

23. Given: ∆ABC  ∆DEF Example 4 Find mF. mEFD + mDEF + mFDE = 180° mEFD + 53 + 90 = 180 mF + 143 = 180 mF = 37° ABC  DEF mABC = mDEF ∆ Sum Thm. Substitute values for mDEF and mFDE. Simplify. Subtract 143 from both sides. Corr. s of  ∆ are . Def. of  s. mDEF = 53° Transitive Prop. of =.

24. Example : Proving Triangles Congruent Given: YWX and YWZ are right angles. YW bisects XYZ. W is the midpoint of XZ. XY  YZ. Prove: ∆XYW  ∆ZYW

25. StatementsReasons 9. Given 7. Reflex. Prop. of  8. Third s Thm.8. X  Z 10. Def. of  ∆ 10. ∆ XYW  ∆ ZYW 6. Def. of mdpt. 5. Given 4. XYW  ZYW 4. Def. of bisector 9. XY  YZ 7. YW  YW 6. XW  ZW 3. Given 3. YW bisects XYZ 2. Rt.   Thm. 2. YWX  YWZ 1. Given 1. YWX and YWZ are rt. s. 5. W is mdpt. of XZ

26. Example Given: AD bisects BE. BE bisects AD. AB  DE, A  D Prove: ∆ABC  ∆DEC

27. 6. Def. of bisector 7. Def. of  ∆s7. ∆ABC  ∆DEC 5. Given 3. ABC  DEC 4. Given 2. BCA  DCE 3. Third s Thm. 2. Vertical s are . 1. Given 1. A  D 4. AB  DE StatementsReasons BE bisects AD 5. AD bisects BE, 6. BC  EC, AC  DC

28. Triangle Congruence: SSS and SAS Holt Geometry Lesson Presentation Lesson Presentation Holt McDougal Geometry

29. For example, you only need to know that two triangles have three pairs of congruent corresponding sides. This can be expressed as the following postulate.

30. Example 1: Using SSS to Prove Triangle Congruence Use SSS to explain why ∆ABC  ∆DBC. It is given that AC  DC and that AB  DB. By the Reflexive Property of Congruence, BC  BC. Therefore ∆ ABC  ∆ DBC by SSS.

31. An included angle is an angle formed by two adjacent sides of a polygon. B is the included angle between sides AB and BC.

33. The letters SAS are written in that order because the congruent angles must be between pairs of congruent corresponding sides. Caution

34. Example: Engineering Application The diagram shows part of the support structure for a tower. Use SAS to explain why ∆ XYZ  ∆ VWZ. It is given that XZ  VZ and that YZ  WZ. By the Vertical s Theorem. XZY  VZW. Therefore ∆ XYZ  ∆ VWZ by SAS.

35. Example Use SAS to explain why ∆ ABC  ∆ DBC. It is given that BA  BD and ABC  DBC. By the Reflexive Property of , BC  BC. So ∆ ABC  ∆ DBC by SAS.

36. Example 4: Proving Triangles Congruent Given: BC ║ AD, BC  AD Prove: ∆ABD  ∆CDB ReasonsStatements 5. SAS Steps 3, 2, 4 5. ∆ ABD  ∆ CDB 4. Reflex. Prop. of  3. Given 2. Alt. Int. s Thm.2. CBD  ADB 1. Given1. BC || AD 3. BC  AD 4. BD  BD

37. Check It Out! Example 4 Given: QP bisects RQS. QR  QS Prove: ∆RQP  ∆SQP ReasonsStatements 5. SAS Steps 1, 3, 4 5. ∆ RQP  ∆ SQP 4. Reflex. Prop. of  1. Given 3. Def. of bisector 3. RQP  SQP 2. Given 2. QP bisects RQS 1. QR  QS 4. QP  QP

38. Practice Given: PN bisects MO, PN  MO Prove: ∆MNP  ∆ONP 1. Given 2. Def. of bisect 3. Reflex. Prop. of  4. Given 5. Def. of  6. Rt.   Thm. 7. SAS Steps 2, 6, 3 1. PN bisects MO 2. MN  ON 3. PN  PN 4. PN  MO 5. PNM and PNO are rt. s 6. PNM  PNO 7. ∆ MNP  ∆ ONP ReasonsStatements

39. Triangle Congruence: ASA, AAS, and HL Holt Geometry Lesson Presentation Lesson Presentation Holt McDougal Geometry

40. An included side is the common side of two consecutive angles in a polygon. The following postulate uses the idea of an included side.

42. Example: Applying ASA Congruence Determine if you can use ASA to prove the triangles congruent. Explain. Two congruent angle pairs are give, but the included sides are not given as congruent. Therefore ASA cannot be used to prove the triangles congruent.

43. Example Determine if you can use ASA to prove NKL  LMN. Explain. By the Alternate Interior Angles Theorem. KLN  MNL. NL  LN by the Reflexive Property. No other congruence relationships can be determined, so ASA cannot be applied.

44. You can use the Third Angles Theorem to prove another congruence relationship based on ASA. This theorem is Angle-Angle-Side (AAS).

46. Example: Using AAS to Prove Triangles Congruent Use AAS to prove the triangles congruent. Given: X  V, YZW  YWZ, XY  VY Prove:  XYZ  VYW

47. Example Use AAS to prove the triangles congruent. Given: JL bisects KLM, K  M Prove: JKL  JML

49. Example: Applying HL Congruence Determine if you can use the HL Congruence Theorem to prove the triangles congruent. If not, tell what else you need to know. According to the diagram, the triangles are right triangles that share one leg. It is given that the hypotenuses are congruent, therefore the triangles are congruent by HL.

50. Example: Applying HL Congruence This conclusion cannot be proved by HL. According to the diagram, the triangles are right triangles and one pair of legs is congruent. You do not know that one hypotenuse is congruent to the other.

51. Example Determine if you can use the HL Congruence Theorem to prove ABC  DCB. If not, tell what else you need to know. Yes; it is given that AC  DB. BC  CB by the Reflexive Property of Congruence. Since ABC and DCB are right angles, ABC and DCB are right triangles. ABC  DCB by HL.

52. Practice Identify the postulate or theorem that proves the triangles congruent. ASA HL SAS or SSS

53. Practice 4. Given: FAB  GED, ABC   DCE, AC  EC Prove: ABC  EDC

54. Lesson Quiz: Part II Continued 5. ASA Steps 3,4 5. ABC  EDC 4. Given 4. ACB  DCE; AC  EC 3.  Supp. Thm.3. BAC  DEC 2. Def. of supp. s 2. BAC is a supp. of FAB; DEC is a supp. of GED. 1. Given 1. FAB  GED ReasonsStatements

55. Triangle Congruence: CPCTC Holt Geometry Lesson Presentation Lesson Presentation Holt McDougal Geometry

56. 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.

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

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

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

60. Example: 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.

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

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

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

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

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

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

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

68. Practice 1. Given: Isosceles ∆PQR, base QR, PA  PB Prove: AR  BQ

69. 4. Reflex. Prop. of 4. P  P 5. SAS Steps 2, 4, 3 5. ∆QPB  ∆RPA 6. CPCTC6. AR = BQ 3. Given3. PA = PB 2. Def. of Isosc. ∆2. PQ = PR 1. Isosc. ∆PQR, base QR Statements 1. Given Reasons Practice Solution

70. Practice 2. Given: X is the midpoint of AC. 1  2 Prove: X is the midpoint of BD.

71. Practice 2 Solution 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 ReasonsStatements 6. DX  BX