1. 5-1 Special Segments in Triangles

2. I. Triangles have four types of special segments:

3. A. Perpendicular bisector Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment every triangle has 3 perpendicular bisectors

5. Examples of perpendicular bisectors In a right triangle, the perpendicular bisectors meet on the triangle. In an acute triangle, the perpendicular bisectors meet inside the triangle. In an obtuse triangle, the perpendicular bisectors meet outside the triangle.

6. Circumcenter The point where all the perp. Bisectors meet in a triangle The point where all the perp. Bisectors meet in a triangle

7. B. Median A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. D

8. PROPERTIES OF MEDIANS every triangle has 3 medians the medians always meet inside the triangle

9. Centroid Point where all the medians meet Point where all the medians meet

10. C. Altitude

11. The three altitudes intersect at G, a point inside the triangle.

12. A B C In right triangle ABC, segment AB and segment BC are two of the altitudes of the triangle

13. Orthocenter Point where all the altitudes meet Point where all the altitudes meet

14. D. Angle bisector

15. - there are 3 angle bisectors in every triangle. -any point on an angle bisector of an angle is equidistant (same distance) from the sides of the angle. -any point on or in the interior of an angle and equidistant from the sides of the angle lies on the angle bisector. -if a triangle is isosceles, the bisector of the vertex angle is also a median and an altitude. -The angle bisectors always meet inside the triangle.

16. Incenter Point where all the angle bisectors meet Point where all the angle bisectors meet

17. 1. Given: < F = 80, < E = 30 DG bisects < EDF Prove: < DGE = 115 F D G E

18. 2.Find x and < 2. MS is an altitude of  MNQ < 1 = 3x + 11 < 2 = 7x + 9 M R Q S N 1 2

19. 3.MS is a median of  MNQ. QS = 3a -14 SN = 2 a + 1 <MSQ = 7a + 1 Find a. Is MS an altitude? M R Q S N 1 2

20. 4. Always, sometimes, never 3 medians intersect in the interior 3 medians intersect in the interior 3 altitudes intersect at a vertex 3 altitudes intersect at a vertex 3 angle bisectors intersect on the exterior 3 angle bisectors intersect on the exterior 3 perpendicular bisectors intersect on the exterior 3 perpendicular bisectors intersect on the exterior

21. 5. Triangle ABC has vertices A(-3, 10) B(9, 2) and C(9, 15) a.Determine the coordinates of P on segment AB so that segment CP is a median of the triangle b.Determine if segment CP is also an altitude of the triangle C A B a.P is the midpoint of segment AB (-3 + 9)/2 (10 + 2)/2 The midpoint is P(3, 6) b.The slope of segment CP is (6 – 15)/(3 – 9) = 3/2 The slope of segment AB is (2 – 10)/(9 - - 3) = -2/3 Since the slopes are opposite reciprocals, CP is perpendicular to AB. Therefore, segment CP is an altitude B

22. 6. Find the indicated information: a. P Q R S Find PQ if segment PS is a median of the triangle QS = x + 5 SR = 3x – 17 And PQ = 2x -8 b. A BC D Find BD if AC is an altitude and m<ACD = 3x + 30, BC = x + 4 and CD = 2x + 8 c. XYZ W Find m<XWZ if m<XWY = 2x – 4 and m<XWZ = 5x – 12. Segment WY is an angle bisector.

23. Solutions to example 6: a.Since segment PS is a median, QS = SR. So, x + 5 = 3x – 17. Then 5 = 2x – 17 22 = 2x 11 = x Finally PQ = 2x – 8 2(11) – 8 PQ = 14 b. If BC is an altitude, then <ACB is a right angle. Then, 3x + 30 = 90. 3x = 60 x = 20 Finally BD = BC + CD x + 4 + 2x + 8 3x + 12 3(20) +12 BD = 72 c. Since segment WY is an angle bisector, it divides the angle in half. So m<XWZ = 2m<XWY. Therefore 5x – 12 = 2(2x - 4) 5x – 12 = 4x - 8 x – 12 = -8 x = 4 So m<XWZ = 5x – 12 5(4) – 12 m<XWZ = 8

24. Example 7: Write at least one conclusion that can be made from each of the following statements. S N E M R L a.Segment SM is an altitude to segment RE b.SN = NE c.M is equidistant from R and E, and <RMS is a right angle d.m<ERN = m<SRN e.segment EL is perpendicular to segment SR Sample answers: a.Segment SM is perpendicular to segment RE <SMR and <SME are right angles b. segment RN is a median and N is the midpoint c. Segment SM is a perpendicular bisector segment SM is a median segment SM is an altitude d.Segment RN is an angle bisector e.Segment EL is an altitude and <ELR and <ELS are right angles

26. 1. The segment that bisects an angle of the  and has one endpoint at a vertex of the  and the other endpoint at another point on the  is called the _________. 1. The segment that bisects an angle of the  and has one endpoint at a vertex of the  and the other endpoint at another point on the  is called the _________.

27. 2. A _________ is a line or segment that passes through the midpoint of a side of a  and is perpendicular to that side. 2. A _________ is a line or segment that passes through the midpoint of a side of a  and is perpendicular to that side.

28. 3. Triangle XYZ has vertices X(-1, 1), 3. Triangle XYZ has vertices X(-1, 1), Y(3, 9), and Z(6, -2). Determine the coordinates of point W on so that is a median of the triangle. Y(3, 9), and Z(6, -2). Determine the coordinates of point W on so that is a median of the triangle.

29. 4. Triangle CPR has vertices C(15, 1), P(9, 11), and R(2, 1). Determine the coordinates of point A on so that is a median of triangle CPR. P(9, 11), and R(2, 1). Determine the coordinates of point A on so that is a median of triangle CPR.

30. 5-2 Inequalities for the sides and angles of a triangle I. Theorem I. Theorem The angle opposite the larger side is always bigger than the angle opposite the shorter side of any triangle The angle opposite the larger side is always bigger than the angle opposite the shorter side of any triangle

31. II. Theorem If one angle is larger than another angle, then the side opposite it is also larger than the side opposite the smaller angle. If one angle is larger than another angle, then the side opposite it is also larger than the side opposite the smaller angle.

32. III. Theorem The perpendicular segment from a point to a line is the shortest segment from the point to the line. The perpendicular segment from a point to a line is the shortest segment from the point to the line. Corollary The perp. Segment from a point to a plane is the shortest segment also. Corollary The perp. Segment from a point to a plane is the shortest segment also.

33. IV.Examples 1. Refer to the figure in example 1 of book. 1. Refer to the figure in example 1 of book. Given: angle A is greater than angle D Given: angle A is greater than angle D Prove: BD is greater than AB Prove: BD is greater than AB

34. 2. Draw triangle JKL with J(-4,2) K (4,3) L (1,-3). List the angles from greatest to least measure. 2. Draw triangle JKL with J(-4,2) K (4,3) L (1,-3). List the angles from greatest to least measure.

35. 5-3 Indirect Proof

36. I. Indirect Proof Steps Assume that the conclusion is false. Assume that the conclusion is false. Show that this leads to a contradiction of a known property, rule, etc… Show that this leads to a contradiction of a known property, rule, etc… Point this out and since it is false, it follows that the conclusion must be true. Point this out and since it is false, it follows that the conclusion must be true. You are proving that the contrapositive of the conditional is true, therefore the original conditional must be true

37. II. A. Exterior Angle Inequality Theorem If an angle is an exterior angle of a triangle, then its measure is greater than the measure of either of its corresponding remote interior angles. If an angle is an exterior angle of a triangle, then its measure is greater than the measure of either of its corresponding remote interior angles. Recall that the sum of the two remote interior angles is equal to the exterior. Since neither of the interior angles are zero, the exterior angle will always be bigger than either of them.

39. B. Definition of an inequality For any real numbers a and b, a > b if and only if there is a positive number c such that a = b + c. For any real numbers a and b, a > b if and only if there is a positive number c such that a = b + c.

40. Properties of Inequalities for Real Numbers p. 254 in book Comparison Property: one of three things has to be true one of three things has to be true a b a=b a b a=b Transitive Property: Transitive Property: If a<b and b<c, then a<c If a<b and b<c, then a<c If a>b and b>c, then a>c If a>b and b>c, then a>c

41. Addition and Subtraction Property: if a>b then a+c>b+c and a-c>b-c if a>b then a+c>b+c and a-c>b-c if a<b then a+c<b+c and a-c<b-c if a<b then a+c<b+c and a-c<b-c Multiplication and Division Property if c>o (positive)and a>b then ac>bc and a/c>b/c if c>o (positive)and a>b then ac>bc and a/c>b/c if c>o (positive)and a o (positive)and a<b then ac<bc and a/c<b/c if c b then ac b then ac<bc and a/c<b/c if c bc and a/c>b/c if c bc and a/c>b/c REMEMBER:if you multiply or divide an inequality by a negative number it switches the direction of the inequality

42. III. Examples 1. Which assumption would you make to start an indirect proof of the statement “two acute angles are congruent”. 1. Which assumption would you make to start an indirect proof of the statement “two acute angles are congruent”.

43. 2. Which assumption would you make to start an indirect proof of the following statements? 2. Which assumption would you make to start an indirect proof of the following statements? Bob took the dog for a walk Bob took the dog for a walk EF is not a perpendicular bisector EF is not a perpendicular bisector 3x = 4 y + 1 3x = 4 y + 1 < 1 is less than or equal to < 2 < 1 is less than or equal to < 2

44. 3. Name the property that justifies if a < b, then a + c < b + c. 3. Name the property that justifies if a < b, then a + c < b + c. 4. Name the property that justifies that if a is less than b, then a cannot be greater than b. 4. Name the property that justifies that if a is less than b, then a cannot be greater than b. 5. Given: 2 y + 8 = 16 5. Given: 2 y + 8 = 16 Prove: y ≠ 5 Prove: y ≠ 5

45. 6. Given: JKL with side lengths 3, 4, 5 6. Given: JKL with side lengths 3, 4, 5 Prove: < K < < L Prove: < K < < L J K L

46. Refer to page 255 in your textbook and try 5 –13 (hint for 8 – 10: use the exterior angle inequality theorem-the exterior angle is greater than the remote interior angles)

47. Answers to p 255-256 5 -13 5. Assume lines l and m do not intersect at x 6. Assume: If the alt int <‘s formed by two parallel lines and a transversal are congruent, then the lines are and a transversal are congruent, then the lines are not parallel. not parallel. 7. Assume Sabrina did not eat the leftover pizza 8. <3, <7, <5, <6 9. <4 and <8 10. <7 because <7 is part of a remote interior angle for the exterior <8 11. Division 12. Addition 13. Transitive

48. 5-4 The Triangle Inequality I. Theorem I. Theorem The sum of the lengths of any two sides of a triangle is greater than the length of the third side. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

49. II. Examples 1. If Mrs. Ewing gave Elizabeth four pieces of tubing measuring 6 m, 7 m, 9 m, and 16 m, how many different triangles could she make? 1. If Mrs. Ewing gave Elizabeth four pieces of tubing measuring 6 m, 7 m, 9 m, and 16 m, how many different triangles could she make?

50. 2. What are the possible lengths for the third side of a triangle with two sides of 8 and 13? 2. What are the possible lengths for the third side of a triangle with two sides of 8 and 13?

51. 3. How many triangles can be made from a rope 10 ft long? 3. How many triangles can be made from a rope 10 ft long?

52. 5-5 Inequalities with two triangles I. Theorem I. Theorem SAS Inequality SAS Inequality

53. II.Theorem SSS Inequality SSS Inequality

54. 5-2 Right Triangles I. What are the postulates for proving 2 right triangles are congruent? I. What are the postulates for proving 2 right triangles are congruent?

55. SAS aka LL LL means Leg Leg

56. ASA, AAS aka LA LA means Leg Angle

57. AAS aka HA HA means Hypotenuse Angle

58. Finally, a new one! HL (only for right triangles/in non-right triangles SSA) HL (only for right triangles/in non-right triangles SSA) HL means Hypotenuse Leg

59. 1. In the figure, is the angle bisector < BAC. Are the triangles congruent? Tell which of the right triangle methods will prove the triangles congruent.

60. 2. Find x so that the right triangles are congruent.

61. 3. What do you need to prove by HA?

62. 4. Name the theorem used to prove the triangles are congruent.

63. 5. Which theorem or postulate states that if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and corresponding acute angle of another right triangle, then the triangles are congruent? 5. Which theorem or postulate states that if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and corresponding acute angle of another right triangle, then the triangles are congruent?

64. Refer to page 248 in your textbook and try problems 6-11

65. Answers to p 248: 6-11 6. Need to know that <B and <D are right angles You already know AC = AC 7. Need to know that ST = TU 8.Need to know that LN = QR or NM = RP 9. 2x + 10 = 5x – 8 so 10 = 3x –8 then 18 = 3x and 6 = x 10. x + 7 = 3x – 5 so 7 = 2x – 5 then 12 = 2x and 6 = x 11. 4x – 26 = 10 then 4x = 36 and x = 9