1. Welcome to Interactive Chalkboard
Glencoe Geometry Interactive Chalkboard
Copyright © by The McGraw-Hill Companies, Inc.
Developed by FSCreations, Inc., Cincinnati, Ohio
Send all inquiries to:
GLENCOE DIVISION
Glencoe/McGraw-Hill
8787 Orion Place
Columbus, Ohio
Welcome to Interactive Chalkboard

2. Splash Screen

3. Lesson 5-1 Bisectors, Medians, and Altitudes
Lesson 5-2 Inequalities and Triangles
Lesson 5-3 Indirect Proof
Lesson 5-4 The Triangle Inequality
Lesson 5-5 Inequalities Involving Two Triangles
Contents

4. Example 1 Use Angle Bisectors Example 2 Segment Measures
Example 3 Use a System of Equations to Find a Point
Lesson 1 Contents

5. Given:
Prove:
Example 1-1a

6. 5. Definition of angle bisector
Proof:
Statements
Reasons 1.
1. Given 2.
2. Angle Sum Theorem 3.
3. Substitution 4.
4. Subtraction Property 5.
5. Definition of angle bisector 6.
6. Angle Sum Theorem 7.
7. Substitution 8.
8. Subtraction Property
Example 1-1a

7. Given: .
Prove:
Example 1-1b

8. Proof: Statements. Reasons
1. Given 2. Angle Sum Theorem 3. Substitution 4. Subtraction Property 5. Definition of angle bisector 6. Angle Sum Theorem 7. Substitution 8. Subtraction Property
1. Given
Example 1-1b

9. ALGEBRA Points U, V, and W are the midpoints of. respectively
ALGEBRA Points U, V, and W are the midpoints of respectively. Find a, b, and c.
Example 1-2a

10. Segment Addition Postulate
Find a.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 14.8 from each side.
Divide each side by 4.
Example 1-2a

11. Segment Addition Postulate
Find b.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 6b from each side.
Subtract 6 from each side.
Divide each side by 3.
Example 1-2a

12. Segment Addition Postulate
Find c.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 30.4 from each side.
Divide each side by 10.
Answer:
Example 1-2a

13. ALGEBRA Points T, H, and G are the midpoints of. respectively
ALGEBRA Points T, H, and G are the midpoints of respectively. Find w, x, and y.
Answer:
Example 1-2b

14. COORDINATE GEOMETRY The vertices of HIJ are H(1, 2), I(–3, –3), and J(–5, 1). Find the coordinates of the orthocenter of HIJ.
Example 1-3a

15. Find an equation of the altitude from The slope of
so the slope of an altitude is
Point-slope form
Distributive Property
Add 1 to each side.
Example 1-3a

16. Next, find an equation of the altitude from I to The
slope of so the slope of an altitude is –6.
Point-slope form
Distributive Property
Subtract 3 from each side.
Example 1-3a

17. Equation of altitude from J
Then, solve a system of equations to find the point of intersection of the altitudes.
Equation of altitude from J
Substitution,
Multiply each side by 5.
Add 105 to each side.
Add 4x to each side.
Divide each side by –26.
Example 1-3a

18. Replace x with in one of the equations to find the y-coordinate.
Rename as improper fractions.
Multiply and simplify.
The coordinates of the orthocenter of
Answer:
Example 1-3a

19. COORDINATE GEOMETRY The vertices of ABC are A(–2, 2), B(4, 4), and C(1, –2). Find the coordinates of the orthocenter of ABC.
Answer: (0, 1)
Example 1-3b

20. End of Lesson 1

21. Example 1 Compare Angle Measures Example 2 Exterior Angles
Example 3 Side-Angle Relationships
Example 4 Angle-Side Relationships
Lesson 2 Contents

22. Inequalities
For any real numbers a and b, a>b if and only if there is a positive number c such that a = b + c.
If 6 = 4 + 2, 6 > 4 and 6 >2.
Theorem 5.8: Exterior Angle Inequality – 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.

23. Determine which angle has the greatest measure.
Explore Compare the measure of 1 to the measures of 2, 3, 4, and 5.
Plan Use properties and theorems of real numbers to compare the angle measures.
Example 2-1a

24. Solve Compare m3 to m1.
By the Exterior Angle Theorem, m1 m3 m4. Since angle measures are positive numbers and from the definition of inequality, m1 > m3.
Compare m4 to m1.
By the Exterior Angle Theorem, m1 m3 m4. By the definition of inequality, m1 > m4.
Compare m5 to m1.
Since all right angles are congruent, 4 5. By the definition of congruent angles, m4 m5. By substitution, m1 > m5.
Example 2-1a

25. Answer: 1 has the greatest measure.
Compare m2 to m5.
By the Exterior Angle Theorem, m5 m2 m3. By the definition of inequality, m5 > m2. Since we know that m1 > m5, by the Transitive Property, m1 > m2.
Examine The results on the previous slides show that m1 > m2, m1 > m3, m1 > m4, and m1 > m5. Therefore, 1 has the greatest measure.
Answer: 1 has the greatest measure.
Example 2-1a

26. Determine which angle has the greatest measure.
Answer: 5 has the greatest measure.
Example 2-1b

27. Use the Exterior Angle Inequality Theorem to list all angles whose measures are less than m14.
By the Exterior Angle Inequality Theorem, m14 > m4, m14 > m11, m14 > m2, and m14 > m4 + m3.
Since 11 and 9 are vertical angles, they have equal measure, so m14 > m9. m9 > m6 and m9 > m7, so m14 > m6 and m14 > m7.
Answer: Thus, the measures of 4, 11, 9,  3,  2, 6, and 7 are all less than m14 .
Example 2-2a

28. Use the Exterior Angle Inequality Theorem to list all angles whose measures are greater than m5.
By the Exterior Angle Inequality Theorem, m10 > m5, and m16 > m10, so m16 > m5, m17 > m5 + m6, m15 > m12, and m12 > m5, so m15 > m5.
Answer: Thus, the measures of 10, 16, 12, 15 and 17 are all greater than m5.
Example 2-2b

29. a. all angles whose measures are less than m4
Use the Exterior Angle Inequality Theorem to list all of the angles that satisfy the stated condition.
a. all angles whose measures are less than m4
b. all angles whose measures are greater than m8
Answer: 5, 2, 8, 7
Answer: 4, 9, 5
Example 2-2c

30. More Inequality Theorems
Theorem 5.9: If one side of a triangle is longer than another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side.
Theorem 5.10: If one angle of a triangle has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.

31. Determine the relationship between the measures of RSU and SUR.
Answer: The side opposite RSU is longer than the side opposite SUR, so mRSU > mSUR.
Example 2-3a

32. Determine the relationship between the measures of TSV and STV.
Answer: The side opposite TSV is shorter than the side opposite STV, so mTSV < mSTV.
Example 2-3b

33. Determine the relationship between the measures of RSV and RUV.
mRSU > mSUR
mUSV > mSUV
mRSU + mUSV > mSUR + mSUV
mRSV > mRUV
Answer: mRSV > mRUV
Example 2-3c

34. Determine the relationship between the measures of the given angles.
a. ABD, DAB
b. AED, EAD
c. EAB, EDB
Answer: ABD > DAB
Answer: AED > EAD
Answer: EAB < EDB
Example 2-3d

35. HAIR ACCESSORIES Ebony is following directions for folding a handkerchief to make a bandana for her hair. After she folds the handkerchief in half, the directions tell her to tie the two smaller angles of the triangle under her hair. If she folds the handkerchief with the dimensions shown, which two ends should she tie?
Example 2-4a

36. Answer: So, Ebony should tie the ends marked Y and Z.
Theorem 5.10 states that if one side of a triangle is longer than another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side. Since X is opposite the longest side it has the greatest measure.
Answer: So, Ebony should tie the ends marked Y and Z.
Example 2-4a

37. KITE ASSEMBLY Tanya is following directions for making a kite
KITE ASSEMBLY Tanya is following directions for making a kite. She has two congruent triangular pieces of fabric that need to be sewn together along their longest side. The directions say to begin sewing the two pieces of fabric together at their smallest angles. At which two angles should she begin sewing?
Answer: A and D
Example 2-4b

38. End of Lesson 2

39. Example 1 Stating Conclusions Example 2 Algebraic Proof
Example 3 Use Indirect Proof
Example 4 Geometry Proof
Lesson 3 Contents

40. Answer: is a perpendicular bisector.
State the assumption you would make to start an indirect proof for the statement is not a perpendicular bisector.
Answer: is a perpendicular bisector.
Example 3-1a

41. State the assumption you would make to start an indirect proof for the statement
Answer:
Example 3-1b

42. If m1 m2 is false, then m1 > m2.
State the assumption you would make to start an indirect proof for the statement m1 is less than or equal to m2.
If m1 m2 is false, then m1 > m2.
Answer: m1 > m2
Example 3-1c

43. Answer: is not congruent to
State the assumption you would make to start an indirect proof for the statement If B is the midpoint of and then is congruent to
The conclusion of the conditional statement is is congruent to The negation of the conclusion is is not congruent to
Answer: is not congruent to
Example 3-1d

44. State the assumption you would make to start an indirect proof of each statement.
a. is not an altitude. b.
Answer: is an altitude.
Answer:
Example 3-1e

45. d. If is an angle bisector of MLP, then MLH is congruent to PLH.
State the assumption you would make to start an indirect proof of each statement.
c. mABC is greater than or equal to mXYZ.
Answer: mABC < mXYZ
Answer: MLH is not congruent to PLH.
Example 3-1e

46. Write an indirect proof.
Given:
Prove:
Indirect Proof:
Step 1 Assume that .
Example 3-2a

47. Step 2 Substitute –2 for y in the equation
Substitution
Multiply.
Add.
This is a contradiction because the denominator cannot be 0.
Example 3-2b

48. Step 3. The assumption leads to a contradiction
Step 3 The assumption leads to a contradiction. Therefore, the assumption that must be false, which means that must be true.
Example 3-2c

49. Write an indirect proof.
Given:
Prove:
Write an indirect proof.
Indirect Proof:
Step 1 Assume that
Example 3-2d

50. Step 2 Substitute –3 for a in the inequality
This is a contradiction because the denominator cannot be 0.
Substitution
Multiply.
Add.
Example 3-2b

51. Step 3. The assumption leads to a contradiction
Step 3 The assumption leads to a contradiction. Therefore, the assumption that must be false, which means that must be true.
Example 3-2b

52. Given: Marta spent less than $156.
CLASSES Marta signed up for three classes at a community college for a little under $156. There was an administration fee of $15, but the class costs varied. How can you show that at least one class cost less than $47?
Answer:
Given: Marta spent less than $156.
Prove: At least one of the classes x cost less than $47. That is,
Example 3-3a

53. Step 1 Assume that none of the classes cost less than $47.
Indirect Proof:
Step 1 Assume that none of the classes cost less than $47.
Step 2 then the minimum total amount Marta spent is However, this is a contradiction since Marta spent less than $156.
Step 3 The assumption leads to a contradiction of a known fact. Therefore, the assumption that must be false. Thus, at least one of the classes cost less than $47.
Example 3-3a

54. Given: David spent less than $135.
SHOPPING David bought four new sweaters for a little under $135. The tax was $7, but the sweater costs varied. How can you show that at least one of the sweaters cost less than $32?
Answer:
Given: David spent less than $135.
Prove: At least one of the sweaters x cost less than $32. That is,
Example 3-3b

55. Step 1 Assume that none of the sweaters cost less than $32.
Step 3 The assumption leads to a contradiction of a known fact. Therefore, the assumption that must be false. Thus, at least one of the sweaters cost less than $32.
Step 1 Assume that none of the sweaters cost less than $32.
Indirect Proof:
Step 2 then the minimum total amount David spent is However, this is a contradiction since David spent less than $135.
Example 3-3b

56. Write an indirect proof.
Given: JKL with side lengths 5, 7, and 8 as shown.
Prove: mK < mL
Example 3-4a

57. Indirect Proof: Step 1 Assume that
Step 2 By angle-side relationships, By substitution, . This inequality is a false statement.
Step 3 Since the assumption leads to a contradiction, the assumption must be false. Therefore, mK < mL.
Example 3-4a

58. Given: ABC with side lengths 8, 10, and 12 as shown.
Prove: mC > mA
Write an indirect proof.
Example 3-4b

59. Indirect Proof: Step 1 Assume that
Step 3 Since the assumption leads to a contradiction, the assumption must be false. Therefore, mC > mA.
Indirect Proof:
Step 1 Assume that
Step 2 By angle-side relationships, By substitution, This inequality is a false statement.
Example 3-4b

60. End of Lesson 3

61. Example 1 Identify Sides of a Triangle
Example 2 Determine Possible Side Length
Example 3 Prove Theorem 5.12
Lesson 4 Contents

62. Theorem 5.11: Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
This theorem will be used to determine whether three segments can form a triangle.

63. Determine whether the measures and
can be lengths of the sides of a triangle.
Answer: Because the sum of two measures is not greater than the length of the third side, the sides cannot form a triangle.
Example 4-1a

64. Determine whether the measures 6. 8, 7. 2, and 5
Determine whether the measures 6.8, 7.2, and 5.1 can be lengths of the sides of a triangle.
Check each inequality.
Answer: All of the inequalities are true, so 6.8, 7.2, and 5.1 can be the lengths of the sides of a triangle.
Example 4-1b

65. Determine whether the given measures can be lengths of the sides of a triangle.
Answer: no
Answer: yes
Example 4-1c

66. Multiple-Choice Test Item In and Which measure cannot be PR?
A 7 B 9 C 11 D 13
Example 4-2a

67. You need to determine which value is not valid.
Read the Test Item
You need to determine which value is not valid.
Solve the Test Item
Solve each inequality to determine the range of values for PR.
Example 4-2a

68. Graph the inequalities on the same number line.
The range of values that fit all three inequalities is
Example 4-2a

69. Examine the answer choices
Examine the answer choices. The only value that does not satisfy the compound inequality is 13 since 13 is greater than Thus, the answer is choice D.
Answer: D
Example 4-2a

70. Multiple-Choice Test Item Which measure cannot be XZ?
A 4 B 9 C 12 D 16
Answer: D
Example 4-2b

71. Perpendicular Segments
Theorem 5.12: The perpendicular segment from a point to a line is the shortest segment from the point to the line.
Corollary 5.1: The perpendicular segment from a point to a plane is the shortest segment from the point to a plane.

72. Given: line through point J Point K lies on t.
Prove: KJ < KH
Example 4-3a

73. 2. Perpendicular lines form right angles.
Proof:
Statements
Reasons 1.
1. Given
are right angles. 2.
2. Perpendicular lines form right angles. 3.
3. All right angles are congruent. 4.
4. Definition of congruent angles 5.
5. Exterior Angle Inequality Theorem 6.
6. Substitution 7.
7. If an angle of a triangle is greater than a second angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.
Example 4-3a

74. Given: is an altitude in ABC.
Prove: AB > AD
Given: is an altitude in ABC.
Example 4-3b

75. Proof: Statements 1. 2. 3. 4. Reasons
Reasons
1. Given 2. Definition of altitude 3. Perpendicular lines form right angles. 4. All right angles are congruent.
is an altitude in
are right angles.
Example 4-3b

76. Proof: Statements 5. 6. 7. 8. Reasons
5. Definition of congruent angles 6. Exterior Angle Inequality Theorem 7. Substitution 8. If an angle of a triangle is greater than a second angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.
Example 4-3b

77. End of Lesson 4

78. Example 1 Use SAS Inequality in a Proof
Example 2 Prove Triangle Relationships
Example 3 Relationships Between Two Triangles
Example 4 Use Triangle Inequalities
Lesson 5 Contents

79. Write a two-column proof.
Given:
Prove:
Example 5-1a

80. 2. Alternate interior angles are congruent.
Proof:
Statements
Reasons 1.
1. Given 2.
2. Alternate interior angles are congruent. 3.
3. Substitution 4.
4. Subtraction Property 5.
5. Given 6.
6. Reflexive Property 7.
7. SAS Inequality
Example 5-1a

81. Given: m1 < m3 E is the midpoint of Write a two-column proof.
Prove: AD < AB
Given: m1 < m3 E is the midpoint of
Write a two-column proof.
Example 5-1b

82. Proof: Statements 1. 2. 3. 4. 5. 6. 7. Reasons
1. Given 2. Definition of midpoint 3. Reflexive Property 4. Given 5. Definition of vertical angles 6. Substitution 7. SAS Inequality
E is the midpoint of
Example 5-1b

83. Given:
Prove:
Example 5-2a

84. Proof: Statements Reasons 1. 1. Given 2. 2. Reflexive Property 3.4.
4. Given 5.
5. Substitution 6.
6. SSS Inequality
Example 5-2a

85. Given: X is the midpoint of MCX is isosceles. CB > CM
Prove:
Example 5-2b

86. Proof: Statements 1. 2. 3. 4. 5. 6. 7. Reasons
Reasons
1. Given 2. Definition of midpoint 3. Given 4. Definition of isosceles triangle 5. Given 6. Substitution 7. SSS Inequality
X is the midpoint of
MCX is isosceles.
Example 5-2b

87. SAS Inequality/Hinge Theorem
If two sides of a triangle are congruent to two sides of another triangle and the included angle in one triangle has a greater measure than the included angle in the other, then the third side of the first triangle is longer than the third side of the second triangle.

88. SSS Inequality
If two sides of a triangle are congruent to two sides of another triangle and the third side in one triangle is longer than the third side in the other, then the angle between the pair of congruent sides in the first triangle is greater than the corresponding angle in the second triangle.

89. The SSS Inequality allows us to conclude that
Write an inequality comparing mLDM and mMDN using the information in the figure.
The SSS Inequality allows us to conclude that
Answer:
Example 5-3a

90. Write an inequality finding the range of values containing a using the information in the figure.
By the SSS Inequality,
Example 5-3b

91. Subtract 15 from each side.
SSS Inequality
Substitution
Subtract 15 from each side.
Divide each side by 9.
Also, recall that the measure of any angle is always greater than 0.
Subtract 15 from each side.
Divide each side by 9.
Example 5-3b

92. The two inequalities can be written as the compound inequality
Answer:
Example 5-3b

93. Write an inequality using the information in the figure. a.
b. Find the range of values containing n.
Answer:
Answer: 6 < n < 25
Example 5-3c

94. HEALTH Doctors use a straight-leg-raising test to determine the amount of pain felt in a person’s back. The patient lies flat on the examining table, and the doctor raises each leg until the patient experiences pain in the back area. Nitan can tolerate the doctor raising his right leg 35° and his left leg 65° from the table. Which foot can Nitan raise higher above the table?
Assume both of Nitan’s legs have the same measurement, the SAS Inequality tells us that the height of the left foot opposite the 65° angle is higher than the height of his right foot opposite the 35° angle. This means that his left foot is raised higher.
Answer: his left foot
Example 5-4a

95. HEALTH Doctors use a straight-leg-raising test to determine the amount of pain felt in a person’s back. The patient lies flat on the examining table, and the doctor raises each leg until the patient experiences pain in the back area. Megan can lift her right foot 18 inches from the table and her left foot 13 inches from the table. Which leg makes the greater angle with the table?
Answer: her right leg
Example 5-4b

96. End of Lesson 5

97. Explore online information about the information introduced in this chapter.
Click on the Connect button to launch your browser and go to the Glencoe Geometry Web site. At this site, you will find extra examples for each lesson in the Student Edition of your textbook. When you finish exploring, exit the browser program to return to this presentation. If you experience difficulty connecting to the Web site, manually launch your Web browser and go to
Online

98. Click the mouse button or press the Space Bar to display the answers.
Transparency 1

99. Transparency 1a

100. Click the mouse button or press the Space Bar to display the answers.
Transparency 2

101. Transparency 2a

102. Click the mouse button or press the Space Bar to display the answers.
Transparency 3

103. Transparency 3a

104. Click the mouse button or press the Space Bar to display the answers.
Transparency 4

105. Transparency 4a

106. Click the mouse button or press the Space Bar to display the answers.
Transparency 5

107. Transparency 5a

108. End of Custom Shows WARNING! Do Not Remove
This slide is intentionally blank and is set to auto-advance to end custom shows and return to the main presentation.
End of Custom Show

109. End of Slide Show