lies in the structural support system of the Ferris Given: In the figure,

Exs. 1–2 In Example 3, we recall the method of construction used to bisect an angle. Although the technique is illustrated, the objective here is to justify the method. EXAMPLE 3 Justify the method for constructing the bisector of an angle. Provide the missing reasons in the proof. GIVEN: by construction by construction See Figure 3.40. PROVE: bisects ∠XYZ YW MW ⬵ NW YM ⬵ YN ∠XYZ PROOF Statements Reasons 1. ; and 1. ? 2. 2. ? 3. 3. ? 4. 4. ? 5. bisects 5. ? ∠XYZ YW ∠MYW ⬵ ∠NYW 䉭YMW ⬵ 䉭YNW YW ⬵ YW MW ⬵ NW YM ⬵ YN ∠XYZ The angle bisector method can be used to construct angles of certain measures. For instance, if a right angle has been constructed, then an angle of measure 45° can be con- structed by bisecting the 90° angle. In Example 4, we construct an angle of measure 30°. EXAMPLE 4 Construct an angle that measures 30°. Solution Figures 3.41a and b: We begin by constructing an equilateral and therefore equiangular triangle. To accomplish this, mark off a line segment of length a. From the endpoints of this line segment, mark off arcs using the same radius length a. The point of intersection determines the third vertex of this triangle, whose angles measure 60° each. Figure 3.41c: By constructing the bisector of one angle, we determine an angle that measures 30°. Y Z M N X W Figure 3.40 a a Figure 3.41 b 60 a a a c 30 쮿 In Example 5, we justify the method for constructing a line perpendicular to a given line from a point not on that line. In the example, point P lies above line ᐉ. 쮿 In Example 6, we recall the method for constructing the line perpendicular to a given line at a point on the line. We illustrate the technique in the example and ask that the student justify the method in Exercise 29. In Example 6, we construct an angle that measures 45°. EXAMPLE 5 GIVEN: P not on 艎 by construction by construction See Figure 3.42. PROVE: Provide the missing statements and reasons in the proof. PQ ⬜ AB AQ ⬵ BQ PA ⬵ PB PROOF Statements Reasons 1. P not on 艎 1. ? and 2. 2. ? 3. 3. ? 4. 4. ? 5. 5. ? 6. 6. ? 7. 7. ? 8. ? 8. If two lines meet to form adjacent , these lines are ⬜ ∠s ⬵ ∠3 ⬵ ∠4 䉭PRA ⬵ 䉭PRB PR ⬵ PR ∠1 ⬵ ∠2 䉭PAQ ⬵ 䉭PBQ PQ ⬵ PQ AQ ⬵ BQ PA ⬵ PB P B Q A R 1 2 3 4 Figure 3.42 EXAMPLE 6 Construct an angle that measures 45°. Solution Figure 3.43a: We begin by constructing a line segment perpendicular to line 艎 at point P. Figure 3.43b: Next we bisect one of the right angles that was determined. The bisector forms an angle whose measure is 45°. 쮿 a P Figure 3.43 b 45 P Exs. 3–5 쮿 As we saw in Example 4, constructing an equilateral triangle is fairly simple. It is also possible to construct other regular polygons, such as a square or a regular hexagon. In the following box, we recall some facts that will help us to perform such constructions. To construct a regular polygon with n sides: 1. Each interior angle must measure degrees; alternatively, each exterior angle must measure degrees. 2. All sides must be congruent. E = 360 n I = n - 2180 n EXAMPLE 7 Construct a regular hexagon having sides of length a. Solution Figure 3.44a: We begin by marking off a line segment of length a. Figure 3.44b: Each exterior angle of the hexagon must measure n = 6 쮿 a a Figure 3.44 120 a a b interior exterior angle 60 a a a c 120 120 60 D E C F B A d Exs. 6–7 ; then each interior angle measures 120°. We construct an equilateral triangle all sides measure a so that a 60° exterior angle is formed. Figure 3.44c: Again marking off an arc of length a for the second side, we construct another exterior angle of measure 60°. Figure 3.44d: This procedure is continued until the regular hexagon ABCDEF is determined. E = 360 6 = 60° Exercises 3.4 In Exercises 1 to 6, use line segments of given lengths a, b, and c to perform the constructions. In Exercises 15 to 18, construct angles having the given measures.

15. 90° and then 45° 16. 60° and then 30°

17. 30° and then 15° 18. 45° and then 105°

HINT: 105 45 60

19. Describe how you would construct an angle

measuring 22.5°.

20. Describe how you would construct an angle

measuring 75°.

21. Construct the complement of the acute

angle shown.

22. Construct the supplement of

the obtuse angle shown. In Exercises 23 to 26, use line segments of lengths a and c as shown.

23. Construct the right triangle with hypotenuse of length

c and a leg of length a. a b c Exercises 1–6

1. Construct a line segment of length 2b. 2. Construct a line segment of length

.

3. Construct a line segment of length . 4. Construct a line segment of length

. 5. Construct a triangle with sides of lengths a, b, and c. 6. Construct an isosceles triangle with a base of length b and legs of length a. In Exercises 7 to 12, use the angles provided to perform the constructions. a - b 1 2 c b + c B A Exercises 7–12

7. Construct an angle that is congruent to acute .

8. Construct an angle that is congruent to obtuse .

9. Construct an angle that has one-half the measure of .

10. Construct an angle that has a measure equal to

.

11. Construct an angle that has twice the measure of .

12. Construct an angle whose measure averages the measures

of and . In Exercises 13 and 14, use the angles and lengths of sides provided to construct the triangle described.

13. Construct the triangle that has sides of lengths r and t with

included angle S. ∠B ∠A ∠A m ∠B - m∠A ∠A ∠B ∠A S R t r Exercises 13, 14

14. Construct the triangle that has a side of length t included

by angles R and S. Q T c a Exercises 23–26

24. Construct an isosceles triangle with base of length c and

altitude of length a. HINT: The altitude lies on the perpendicular bisector of the base.

25. Construct an isosceles triangle with a vertex angle of 30°

and each leg of length c.

26. Construct a right triangle with base angles of 45° and

hypotenuse of length c. In Exercises 27 and 28, use the given angle and the line segment of length b.

27. Construct the right triangle in which acute angle R has a

side one leg of the triangle of length b. b R Exercises 27, 28 Video exercises are available on DVD.

28. Construct an isosceles triangle with base of length b and

congruent base angles having the measure of angle R. See the figure for Exercise 27.

29. Complete the justification of the construction of the line

perpendicular to a given line at a point on that line. Given: Line m, with point P on m by construction by construction Prove:

30. Complete the justification of the construction of the

perpendicular bisector of a line segment. Given: with by construction Prove: and

31. To construct a regular hexagon, what

measure would be necessary for each interior angle? Construct an angle of that measure.

32. To construct a regular octagon, what measure would be

necessary for each interior angle? Construct an angle of that measure.

33. To construct a regular dodecagon 12 sides, what

measure would be necessary for each interior angle? Construct an angle of that measure.

34. Draw an acute triangle and construct the three medians of

the triangle. Do the medians appear to meet at a common point? Í CD ⬜ AB AM ⬵ MB AC ⬵ BC ⬵ AD ⬵ BD AB Í SP ⬜ m QS ⬵ RS PQ ⬵ PR

35. Draw an obtuse triangle and construct the three altitudes

of the triangle. Do the altitudes appear to meet at a common point? HINT: In the construction of two of the altitudes, sides need to be extended.

36. Draw a right triangle and construct the angle bisectors of

the triangle. Do the angle bisectors appear to meet at a common point?

37. Draw an obtuse triangle and construct the three

perpendicular bisectors of its sides. Do the perpendicular bisectors of the three sides appear to meet at a common point?

38. Construct an equilateral triangle and its three altitudes.

What does intuition tell you about the three medians, the three angle bisectors, and the three perpendicular bisectors of the sides of that triangle?

39. A carpenter has placed a

square over an angle in such a manner that and see drawing. What can you conclude about the location of point D?

40. In right triangle ABC,

. Also, BC = a, CA = b, and AB = c. Construct the bisector of so that it intersects at point D. Now construct perpendicular to and with E on . In terms of a, b, and c, find the length of . EA AB AB DE CA ∠B m ∠C = 90° BD ⬵ CD AB ⬵ AC Q m R S P M B A D C A D B C Lemma Inequality of Sides and Angles in a Triangle The Triangle Inequality Inequalities in a Triangle 3.5 KEY CONCEPTS Important inequality relationships exist among the measured parts of a triangle. To establish some of these, we recall and apply some facts from both algebra and geometry. A more in-depth review of inequalities can be found in Appendix A, Section A.3. Let a and b be real numbers. read “a is greater than b” if and only if there is a positive number p for which . a = b + p a 7 b DEFINITION For instance, , because there is the positive number 5 for which . Because , we also know that and . In geometry, let A-B-C on so that ; then , because BC is a positive number. AC 7 AB AB + BC = AC AC 7 7 5 7 7 2 5 + 2 = 7 9 = 4 + 5 9 7 4 Exs. 1–3 LEMMAS HELPING THEOREMS We will use the following theorems to help us prove the theorems found later in this sec- tion. In their role as “helping” theorems, each of the five boxed statements that follow is called a lemma. We will prove the first four lemmas, because their content is geometric. A B C Figure 3.45 B C D A 1 2 Figure 3.46 1 2 3 Figure 3.47 C A B Figure 3.48 If B is between A and C on , then and . The measure of a line segment is greater than the measure of any of its parts. See Figure 3.45. AC 7 BC AC 7 AB AC LEMMA 3.5.1 PROOF By the Segment-Addition Postulate, . According to the Ruler Postulate, meaning BC is positive; it follows that . Similarly, . These relationships follow directly from the definition of . a 7 b AC 7 BC AC 7 AB BC 7 AC = AB + BC If separates into two parts and , then and . The measure of an angle is greater than the measure of any of its parts. See Figure 3.46. m ∠ABC 7 m∠2 m ∠ABC 7 m∠1 ∠2 ∠1 ∠ABC BD LEMMA 3.5.2 PROOF By the Angle-Addition Postulate, . Using the Protractor Postulate, ; it follows that . Similarly, . m ∠ABC 7 m∠2 m ∠ABC 7 m∠1 m ∠2 7 0 m ∠ABC = m∠1 + m∠2 If is an exterior angle of a triangle and and are the nonadjacent interior angles, then and . The measure of an exterior angle of a trian- gle is greater than the measure of either nonadjacent interior angle. See Figure 3.47. m ∠3 7 m∠2 m ∠3 7 m∠1 ∠2 ∠1 ∠3 LEMMA 3.5.3 PROOF Because the measure of an exterior angle of a triangle equals the sum of measures of the two nonadjacent interior angles, . It follows that and . m ∠3 7 m∠2 m ∠3 7 m∠1 m ∠3 = m∠1 + m∠2 In , if is a right angle or an obtuse angle, then and . If a triangle contains a right or an obtuse angle, then the measure of this angle is greater than the measure of either of the remaining angles. See Figure 3.48. m ∠C 7 m∠B m ∠C 7 m∠A ∠C 䉭ABC LEMMA 3.5.4 PROOF In , . With being a right angle or an obtuse angle, ; it follows that . Then each angle and must be acute. Thus, and . The following theorem also a lemma is used in Example 1. Its proof not given depends on the definition of “is greater than,” which is found on the previous page. m ∠C 7 m∠B m ∠C 7 m∠A ∠B ∠A m ∠A + m∠B … 90° m ∠C Ú 90° ∠C m ∠A + m∠B + m∠C = 180° 䉭ABC If and , then . a + c 7 b + d c 7 d a 7 b LEMMA 3.5.5 Addition Property of Inequality