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15. 84 ° 30. 48° 16. 13 ¾ ° 31. 48° 17. (90 – 2x) ° 32. 42°

15. 84 ° 30. 48° 16. 13 ¾ ° 31. 48° 17. (90 – 2x) ° 32. 42° 18. 33.2 ° 33. 120°; 360° 19. 162 ° 34. 37.5° 20. 61 ° 35. 18° 21. 48 °; 48° 39. Measures of ext s will be sum of pairs of

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15. 84 ° 30. 48° 16. 13 ¾ ° 31. 48° 17. (90 – 2x) ° 32. 42°

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  1. 15. 84° 30. 48° 16. 13 ¾ ° 31. 48° 17. (90 – 2x)° 32. 42° 18. 33.2° 33. 120°; 360° 19. 162° 34. 37.5° 20. 61° 35. 18° 21. 48°; 48° 39. Measures of ext s will be sum of pairs of 22. 128°; 128° remote int s: 155°, 65°, and 140° 23. 15°; 60°; 105° 54. isos 24. a. Given 55. scalene b. mF = 90° 56. scalene c. ΔSum Thm. 57. ΔACD is equilateral d. Subst. e. mD + mE = 90° f. Def. of comp s 29. 36°

  2. FG, GH, FH, F, G, H Warm Up 1.Name all sides and angles of ∆FGH. 2. What is true about K and L? Why? 3.What does it mean for two segments to be congruent?  ;Third s Thm. They have the same length.

  3. 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.

  4. Helpful Hint Two vertices that are the endpoints of a side are called consecutive vertices. For example, P and Q are consecutive vertices. To name a polygon, write the vertices in consecutive order. For example, you can name polygon PQRS as QRSP or SRQP, but not as PRQS. In a congruence statement, the order of the vertices indicates the corresponding parts.

  5. Sides: LM EF, MN  FG, NP  GH, LP  EH Helpful Hint When you write a statement such as ABCDEF, you are also stating which parts are congruent. Example 1 If polygon LMNP polygon EFGH, identify all pairs of corresponding congruent parts. Angles: L  E, M  F, N  G, P  H

  6. Example 2A: 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.

  7. 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 =

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

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