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Angle Relationships in Triangles

Angle Relationships in Triangles. Holt Geometry. Lesson Presentation. Holt McDougal Geometry. Review Topics. Vertical Angles Parallel Lines cut by a Transversal Linear Angles (Straight Angle) Complementary and Supplementary Angles Reflexive, Symmetric and Transitive Properties Congruence

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Angle Relationships in Triangles

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  1. Angle Relationships in Triangles Holt Geometry 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. Exterior Interior

  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. 4 is an exterior angle. Exterior Interior 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. 4 is an exterior angle. Exterior The remote interior angles of 4 are 1 and 2. Interior 3 is an interior angle.

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

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

  10. Example 4 Find mP and mT. P  T Third s Thm. mP = mT Def. of s. 2x2= 4x2 – 32 Substitute 2x2 for mP and 4x2 – 32 for mT. –2x2 = –32 Subtract 4x2 from both sides. x2 = 16 Divide both sides by -2. So mP = 2x2 = 2(16) = 32°. Since mP = mT,mT =32°.

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

  12. Congruent Triangles Holt Geometry Lesson Presentation Holt McDougal Geometry

  13. 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.

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

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

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

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

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

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

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

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

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

  23. Example Given:ADbisectsBE. BEbisectsAD. ABDE, A D Prove:∆ABC  ∆DEC

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

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

  26. 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.

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

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

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

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

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

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

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

  34. Statements Reasons 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 1. Given 2. Def. of bisect 3. Reflex. Prop. of  4. Given 5. Def. of  6. Rt.   Thm. 7. SAS Steps 2, 6, 3 Practice Given: PN bisects MO,PN  MO Prove: ∆MNP  ∆ONP

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

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

  37. 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.

  38. 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. Example Determine if you can use ASA to prove NKL LMN. Explain.

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

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

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

  42. 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.

  43. 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.

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