In right triangle ABC,

The paragraph proof in Example 1 could have been written in this standard format. EXAMPLE 1 Give a paragraph proof for the following problem. See Figure 3.49. GIVEN: and PROVE: PROOF: If and , then by Lemma 3.5.5. But and by the Segment-Addition Postulate. Using substitution, it follows that . AC 7 CE CD + DE = CE AB + BC = AC AB + BC 7 CD + DE BC 7 DE AB 7 CD AC 7 CE BC 7 DE AB 7 CD A B C D E Figure 3.49 Geometry in the Real World A carpenter’s “plumb” determines the shortest distance to a horizontal line. A vertical brace provides structural support for the roof. PROOF Statements Reasons 1. and 1. Given 2. 2. Lemma 3.5.5 3. and 3. Segment-Addition Postulate 4. 4. Substitution AC 7 CE CD + DE = CE AB + BC = AC AB + BC 7 CD + DE BC 7 DE AB 7 CD The paragraph proof and the two-column proof of Example 1 are equivalent. In either format, statements must be ordered and justified. The remaining theorems are the “heart” of this section. Before studying the theorem and its proof, it is a good idea to visualize each theorem. Many statements of inequality are intuitive; that is, they are easy to believe even though they may not be easily proved. Study Theorem 3.5.6 and consider Figure 3.50, in which it appears that . m ∠C 7 m∠B Exs. 4–8 If one side of a triangle is longer than a second side, then the measure of the angle oppo- site the longer side is greater than the measure of the angle opposite the shorter side. THEOREM 3.5.6 B 4 6 A C Figure 3.50 EXAMPLE 2 Provide a paragraph proof of Theorem 3.5.6. GIVEN: , with [See Figure 3.51a.] PROVE: PROOF: Given with , we use the Ruler Postulate to locate point D on so that in Figure 3.51b. Now in the isosceles triangle BDC. By Lemma 3.5.2, ; therefore, by substitution. By Lemma 3.5.3, because is an exterior angle of . Using the two starred statements, we can conclude by the Transitive Property of Inequality that ; that is, in Figure 3.51a. m ∠B 7 m∠A m ∠ABC 7 m∠A 䉭ADB ∠5 m ∠5 7 m∠A m ∠ABC 7 m∠5 m ∠ABC 7 m∠2 m ∠2 = m∠5 CD ⬵ BC AC AC 7 BC 䉭ABC m ∠B 7 m∠A AC 7 BC 䉭ABC B A C a Figure 3.51 B A C b 3 5 4 2 1 D 쮿 쮿 The relationship described in Theorem 3.5.6 extends, of course, to all sides and all angles of a triangle. That is, the largest of the three angles of a triangle is opposite the longest side, and the smallest angle is opposite the shortest side. Technology Exploration Use computer software if available. 1. Draw a with as the longest side. 2. Measure A, B, and C. 3. Show that C has the greatest measure. ∠ ∠ ∠ ∠ AB 䉭 ABC A C B 80 40 Figure 3.52 C B A Figure 3.53 EXAMPLE 3 Given that the three sides of not shown are , , and , arrange the angles by size. Solution Because , the largest angle is , which lies opposite . The angle intermediate in size is , which lies opposite . The smallest angle is , which lies opposite the shortest side, . Thus, the order of the angles by size is 쮿 m ∠B 7 m∠A 7 m∠C AB ∠C BC ∠A AC ∠B AC 7 BC 7 AB AC = 6 BC = 5 AB = 4 䉭ABC The converse of Theorem 3.5.6 is also true. It is necessary, however, to use an indirect proof to establish the converse. Recall that this method of proof begins by sup- posing the opposite of what we want to show. Because this assumption leads to a con- tradiction, the assumption must be false and the desired claim is therefore true. Study Theorem 3.5.7 and consider Figure 3.52, in which and . It appears that the longer side lies opposite the larger angle; that is, it appears that . BC 7 AC m ∠B = 40° m ∠A = 80° Discover Using construction paper and a protractor, draw so that , , and . Measure the length of each side. a Which side is longest? b Which side is shortest? ANSWERS m ∠T = 45° m ∠S = 60° m ∠R = 75° 䉭 RST a b RS ST If the measure of one angle of a triangle is greater than the measure of a second angle, then the side opposite the larger angle is longer than the side opposite the smaller angle. THEOREM 3.5.7 The proof of Theorem 3.5.7 depends on this fact: Given real numbers a and b, only one of the following can be true. a 7 b , a = b , or a 6 b EXAMPLE 4 Prove Theorem 3.5.7 by using an indirect approach. GIVEN: with See Figure 3.53. PROVE: PROOF: Given with , assume that . But if , then , which contradicts the hypothesis. Also, if , then it follows by Theorem 3.5.6 that , which also contradicts the hypothesis. Thus, the assumed statement must be false, and it follows that . AC 7 BC m ∠B 6 m∠A AC 6 BC m ∠B = m∠A AC = BC AC … BC m ∠B 7 m∠A 䉭ABC AC 7 BC m ∠B 7 m∠A 䉭ABC EXAMPLE 5 Given not shown in which and , write an extended inequality that compares the lengths of the three sides. Solution Because the sum of angles of is 180°, it follows that . With , it follows that the sides opposite these are unequal in the same order. That is, 쮿 ST 7 RT 7 SR ∠s m ∠R 7 m∠S 7 m∠T m ∠T = 45° 䉭RST m ∠S = 55° m ∠R = 80° 䉭RST 쮿 The following corollary is a consequence of Theorem 3.5.7. D E F G P Figure 3.54 Exs. 9–12 The perpendicular line segment from a point to a line is the shortest line segment that can be drawn from the point to the line. COROLLARY 3.5.8 In Figure 3.54, , , and . In every case, is oppo- site an acute angle of a triangle, whereas the second segment is always opposite a right angle necessarily the largest angle of the triangle involved. With we say that PD is the distance from P to 艎. Corollary 3.5.8 can easily be extended to three dimensions. PD ⬜ , PD PD 6 PG PD 6 PF PD 6 PE The perpendicular line segment from a point to a plane is the shortest line segment that can be drawn from the point to the plane. COROLLARY 3.5.9 In Figure 3.55, is a leg of each right triangle shown. With the hypotenuse of , the hypotenuse of , and the hypotenuse of , the length of is less than that of , , , or any other line segment joining point P to a point in plane R. The length of is known as the distance from point P to plane R. Our final theorem shows that no side of a triangle can have a length greater than or equal to the sum of the lengths of the other two sides. In the proof, the relationship is validated for only one of three possible inequalities. Theorem 3.5.10 is often called the Triangle Inequality. See Figure 3.56. PD PG PF PE PD 䉭PDG PG 䉭PDF PF 䉭PDE PE PD E F G D P R Figure 3.55 B C D A Figure 3.56 Exs. 13–14 The sum of the lengths of any two sides of a triangle is greater than the length of the third side. THEOREM 3.5.10 왘 Triangle Inequality GIVEN: PROVE: PROOF: Draw . Because the shortest segment from a point to is the perpendicular segment, and . Using Lemma 3.5.5, we add the inequalities; . By the Segment- Addition Postulate, the sum can be replaced by BC to yield . The following statement is an alternative and expanded form of Theorem 3.5.10. If a, b, and c are the lengths of the sides of a triangle and c is the length of any side, then . a - b 6 c 6 a + b BA + CA 7 BC BD + CD BA + CA 7 BD + CD CA 7 CD BA 7 BD AD AD ⬜ BC BA + CA 7 BC 䉭ABC The length of any side of a triangle must lie between the sum and difference of the lengths of the other two sides. THEOREM 3.5.10 왘 Triangle Inequality From Example 6, you can see that the length of one side cannot be greater than or equal to the sum of the lengths of the other two sides. Considering the alternative form of Theorem 3.5.10, we see that in part a. When 5 [as in part a] is replaced by 7 [as in part b] or 8 [as in part c], this inequality becomes a false statement. Part d of Example 6 shows that the length of the third side must be between 1 and 7. Our final example illustrates a practical application of inequality relationships in triangles. 4 - 3 6 5 6 4 + 3 EXAMPLE 6 Can a triangle have sides of the following lengths? a 3, 4, and 5 b 3, 4, and 7 c 3, 4, and 8 d 3, 4, and x Solution a Yes, because no side has a length greater than or equal to the sum of the lengths of the other two sides that is, b No, because need c No, because need d Yes, if 쮿 4 - 3 6 x 6 4 + 3 4 - 3 6 8 6 3 + 4 8 7 3 + 4 4 - 3 6 7 6 3 + 4 7 = 3 + 4 4 - 3 6 5 6 3 + 4 EXAMPLE 7 On a map, firefighters are located at points A and B. A fire has broken out at point C. Which group of firefighters is nearer the location of the fire? See Figure 3.57. Solution With and , the side opposite has a greater length than the side opposite . It follows that . Because the distance from B to C is less than the distance from A to C, the firefighters at site B should be dispatched to the fire located at C. NOTE: In Example 7 we assume that highways from A and B to C are equally accessible. 쮿 AC 7 BC ∠A ∠B m ∠B = 46° m ∠A = 43° Exs. 15–18 A C B 43 46 Figure 3.57 Exercises 3.5 In Exercises 1 to 10, classify each statement as true or false.

1. is the longest side of

. 䉭ABC AB

8. If is the bisector of

, then .

9. DA 7 AC DG 7 DE

∠EDF DG A B C 65° 70° 45° Exercises 1, 2

2. 3. DB 7 AB

AB 6 BC A B 50° C D 100° 110° 100° Exercises 3, 4

4. Because , it follows that

.

5. m

∠A + m∠B = m∠C DA = DC m ∠A = m∠B C B A 3 5 4 Exercises 5, 6 6. 7. DF 7 DE + EF m ∠A 7 m∠B E F G D 60° 50° Exercises 7, 8 45° 35° 10° 10° D E C B A Exercises 9, 10

10. 11. If possible, draw a triangle whose angles measure:

a 100°, 100°, and 60° b 45°, 45°, and 90°

12. If possible, draw a triangle whose angles measure:

a 80°, 80°, and 50° b 50°, 50°, and 80°

13. If possible, draw a triangle whose sides measure:

a 8, 9, and 10 b 8, 9, and 17 c 8, 9, and 18

14. If possible, draw a triangle whose sides measure:

a 7, 7, and 14 b 6, 7, and 14 c 6, 7, and 8 In Exercises 15 to 18, describe the triangle , not shown as scalene, isosceles, or equilateral. Also, is the triangle acute, right, or obtuse?

15. and

16. and .

17. 18.

and

19. Two of the sides of an isosceles triangle have lengths

of 10 cm and 4 cm. Which length must be the length of the base?

20. The sides of a right triangle have lengths of 6 cm, 8 cm,

and 10 cm. Which length is that of the hypotenuse?

21. A triangle is both isosceles and acute. If one angle of the

triangle measures 36°, what is the measure of the largest angles of the triangle? What is the measure of the smallest angles of the triangle?

22. One of the angles of an isosceles triangle measures 96°.

What is the measure of the largest angles of the triangle? What is the measure of the smallest angles of the triangle? m ∠Y = 40° m ∠X = 70° m ∠X = m∠Y = 40° ∠Y ⬵ ∠Z m ∠X = 60° m ∠Y = 47° m ∠X = 43° 䉭XYZ CE = ED

23. NASA in Huntsville, Alabama at point H, has called a

manufacturer for parts needed as soon as possible. NASA will, in fact, send a courier for the necessary equipment. The manufacturer has two distribution centers located in nearby Tennessee—one in Nashville at point N and the other in Jackson at point J. Using the angle measurements indicated on the accompanying map, determine to which town the courier should be dispatched to obtain the needed parts.

26. Given: Equilateral and

D-B-C Prove: DA 7 AC 䉭ABC N H J 44° Tennessee Alabama 72° W T S 54° 42°

24. A tornado has just struck a small Kansas community at

point T. There are Red Cross units stationed in both Salina at point S and Wichita at point W. Using the angle measurements indicated on the accompanying map, determine which Red Cross unit would reach the victims first. Assume that both units have the same mode of travel and accessible roadways available. In Exercises 25 and 26, complete each proof.

25. Given:

Prove: m ∠ABD 7 m∠DBF m ∠CBD 7 m∠EBF m ∠ABC 7 m∠DBE B F E D C A PROOF Statements Reasons 1. ? 1. Given 2. 2. Addition Property of Inequality 3. 3. ? 4. ? 4. Substitution m ∠DBE + m∠EBF m ∠CBD and m∠DBF = m ∠ABD = m∠ABC + m ∠DBE + m∠EBF m ∠ABC + m∠CBD 7 D C B A PROOF Statements Reasons 1. ? 1. Given 2. is equiangular, so 2. ? 3. 3. The measure of an ext. of of a is greater than the measure of either nonadjacent int. 4. ? 4. Substitution 5. ? 5. ? ∠ 䉭 ∠ 䉭ABD ∠D m ∠ABC 7 m∠D m ∠ABC = m∠C 䉭ABC In Exercises 27 and 28, construct proofs.

27. Given: Quadrilateral RSTU with diagonal

and are right Prove: TS 7 UR ∠s ∠TUS ∠R US R S U T

28. Given: Quadrilateral ABCD with

Prove: DC 7 AB AB ⬵ DE D C A E B

29. For and

not shown, suppose that and but that . Draw a conclusion regarding the lengths of and .

30. In not shown, point Q lies on

so that bisects . If , draw a conclusion about the relative lengths of and . In Exercises 31 to 34, apply a form of Theorem 3.5.10.

31. The sides of a triangle have lengths of 4, 6, and x. Write

an inequality that states the possible values of x.

32. The sides of a triangle have lengths of 7, 13, and x. As in

Exercise 31, write an inequality that describes the possible values of x.

33. If the lengths of two sides of a triangle are represented by

and in which x is positive, describe in terms of x the possible lengths of the third side whose length is represented by y. 3x + 7 2x + 5 QP NQ MN 6 MP ∠NMP MQ NP 䉭MNP EF BC m ∠A 6 m∠D AB ⬵ DE AC ⬵ DF 䉭DEF 䉭ABC

34. Prove by the indirect method: “The length of a diagonal of

a square is not equal in length to the length of any of the sides of the square.”

35. Prove by the indirect method:

Given: is not isosceles Prove:

36. Prove by the indirect method:

Given: Scalene in which bisects point W lies on . Prove : is not perpendicular to . In Exercises 37 and 38, prove each theorem.

37. The length of the median from the vertex of an isosceles

triangle is less than the length of either of the legs.

38. The length of an altitude of an acute triangle is less than

the length of either side containing the same vertex as the altitude. XY ZW XY ∠XYZ ZW 䉭XYZ PM Z PN 䉭MPN