In right triangle XYZ, and . Where Diagonal separates pentagon

Two properties that were introduced earlier Property 3 of Section 5.1 are now recalled. The subtraction operation of the property is needed for the proof of Theorem 5.6.1. If a b = c d , then a ; b b = c ; d d If a line is parallel to one side of a triangle and intersects the other two sides, then it di- vides these sides proportionally. THEOREM 5.6.1 GIVEN: In Figure 5.48, with and with intersecting at D and at E PROVE: PROOF: Because , . With as a common angle for and , it follows by AA that these triangles are similar. Now by CSSTP By Property 3 of Section 5.1, Because and , the proportion becomes Using Property 2 of Section 5.1, we can invert both fractions to obtain the desired conclusion: AD DB = AE EC DB AD = EC AE AC - AE = EC AB - AD = DB AB - AD AD = AC - AE AE AB AD = AC AE 䉭ABC 䉭ADE ∠A ∠1 ⬵ ∠2 Í DE 7 BC AD DB = AE EC AC AB Í DE Í DE 7 BC 䉭ABC C B D A E 1 2 Figure 5.48 Exs. 3–6 When three or more parallel lines are cut by a pair of transversals, the transversals are divided proportionally by the parallel lines. COROLLARY 5.6.2 GIVEN: in Figure 5.49 on page 262 PROVE: AB BC = DE EF p 1 7 p 2 7 p 3 쮿 PICTURE PROOF OF COROLLARY 5.6.2 p 1 A p 2 p 3 D B G E C F t 1 t 2 Figure 5.49 p 1 A p 2 p 3 F B G E C D t 1 t 2 H p 4 Figure 5.50 In Figure 5.49, draw as an auxiliary line segment. On the basis of Theorem 5.6.1, we see that in and that in . By the Transitive Property of Equality, AB BC = DE EF . 䉭ADF AG GF = DE EF 䉭ACF AB BC = AG GF AF NOTE: By interchanging the means, we can write the last proportion in the form . AB DE = BC EF EXAMPLE 3 Given parallel lines , , , and cut by and so that , , , and , find FG and CD. See Figure 5.50. Solution Because the transversals are divided proportionally, so Then FG = 3 2 = 1 1 2 and CD =

20 3

= 6 2 3 4 FG =

6 and

3 CD = 20 4 3 = 2 FG = CD 5 AB EF = BC FG = CD GH GH = 5 BC = 2 EF = 3 AB = 4 t 2 t 1 p 4 p 3 p 2 p 1 쮿 Exs. 7, 8 The following activity leads us to the relationship described in Theorem 5.6.3. Discover On a piece of paper, draw or construct whose sides measure , , and . Then construct the angle bisector of . How does compare to ? ANSWER BC DC AB AD ∠B BD AC = 5 BC = 6 AB = 4 䉭 ABC Though not by chance, it may come as a surprise that and . It seems that the bisector of an angle included by two sides of a triangle separates the third side into segments whose lengths are proportional to the lengths of the two sides forming the angle. AB BC = AD DC a 4 6 = 2 3 b AB AD = BC DC athat is, 4 2 = 6 3 b C A B a C A D B b The proof of Theorem 5.6.3 requires the use of Theorem 5.6.1. If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle. THEOREM 5.6.3 왘 The Angle-Bisector Theorem b A D C B 1 2 E 3 4 a A D C B 1 2 Figure 5.51 GIVEN: in Figure 5.51a, in which bisects PROVE: PROOF: We begin by extending beyond C there is only one line through B and C to meet the line drawn through A parallel to . [See Figure 5.51b.] Let E be the point of intersection. These lines must intersect; otherwise, would have two parallels, and , through point C. Because , we have by Theorem 5.6.1. Now because bisects , corresponding angles for parallel lines, and alternate interior angles for parallel lines. By the Transitive Property, , so is isosceles with . Using substitution, the starred proportion becomes by inversion The “Prove statement” of the preceding theorem indicates that one form of the pro- portion described is given by comparing lengths as shown: Equivalently, the proportion could compare lengths like this: Other forms of the proportion are also possible segment at left segment at right = side at left side at right segment at left side at left = segment at right side at right AC AD = CB DB or AD AC = DB CB EC ⬵ AC 䉭ACE ∠3 ⬵ ∠4 ∠2 ⬵ ∠4 ∠1 ⬵ ∠3 ∠ACB CD ∠1 ⬵ ∠2 EC AD = CB DB CD 7 EA CD BC AE DC BC AD AC = DB CB ∠ACB CD 䉭ABC EXAMPLE 4 For in Figure 5.52, and . If bisects and , find XZ. Solution Let . We know that , so . Therefore, Then . Because , we have . XZ = 2 + 3 1 3 = 5 1 3 XZ = XW + WZ WZ = 3 1 3 x = 10 3 = 3 1 3 3x = 10 3 2 = 5 x YX XW = YZ WZ WZ = x XW = 2 ∠XYZ YW YZ = 5 XY = 3 䉭XYZ Y X W Z Figure 5.52 쮿 Exs. 9–13 쮿 In the following example, we provide an alternative solution to a problem of the type found in Example 5. EXAMPLE 5 In Figure 5.52 shown in Example 6, suppose that has sides of lengths , , and . If bisects , find XW and WZ. Solution Let ; then , and becomes . From this proportion, we can find y as follows. Then and . WZ = 5 - 2 1 7 = 2 6 7 XW = 15 7 = 2 1 7 y = 15 7 15 = 7y 15 - 3y = 4y 35 - y = 4y

3 4

= y 5 - y XY YZ = XW WZ WZ = 5 - y XW = y ∠XYZ YW XZ = 5 YZ = 4 XY = 3 䉭XYZ 쮿 EXAMPLE 6 In Figure 5.52, is isosceles with If and , find XW and WZ. Solution Because the ratio XY:YZ is 3:6, or 1:2, the ratio XW:WZ is also 1:2. Thus, we can represent these lengths by With in the isosceles triangle, the statement becomes , so , and . Now and . WZ = 4 XW = 2 a = 2 3a = 6 a + 2a = 6 XW + WZ = XZ XZ = 6 XW = a and WZ = 2a YZ = 6 XY = 3 XZ ⬵ YZ. 䉭XYZ 쮿 Y X W Z Figure 5.52 You will find the proof of the following theorem in the Perspective on History sec- tion at the end of this chapter. In Ceva’s Theorem, point D is any point in the interior of the triangle. See Figure 5.53a. The auxiliary lines needed to complete the proof of Ceva’s Theorem are shown in Figure 5.53b. In the figure, line 艎 is drawn through vertex C so that it is parallel to Then and are extended to meet 艎 at R and S, respectively. AF BE AB . Let point D be any point in the interior of , and let , , and be the line segments determined by D and vertices of . Then the product of the ratios of the lengths of the segments of each of the three sides taken in order from a given vertex of the triangle equals 1; that is, AG GB BF FC CE EA = 1 䉭ABC CG AF BE 䉭ABC THEOREM 5.6.4 왘 Ceva’s Theorem For Figure 5.53, Ceva’s Theorem can be stated in many equivalent forms: In each case, we select a vertex and form ratios of the lengths of segments of sides in a set order. We will apply Ceva’s Theorem in Example 7. AE EC CF FB BG GA = 1, CF FB BG GA AE EC = 1, etc. G a E F B A C D Figure 5.53 b G E F B A C D R S EXAMPLE 7 In with interior point D, , , , , and . Find TK. See Figure 5.54. Solution Let . Applying Ceva’s Theorem and following a counterclockwise path beginning at vertex R, we have Then and so becomes Then and ; thus, . TK = 2.5 a = 2.5 2a = 5 2a 5 = 1. 6 2 4 1 4 1 3 1 a 5 = 1

6 4

4 3 a 5 = 1 RG GS SH HT TK KR = 1. TK = a KR = 5 HT = 3 SH = 4 GS = 4 RG = 6 䉭RST G S H D K R T Figure 5.54 쮿 Ex. 14

5. Given: ,

, , , Find: EF , FG, GH See the figure for Exercise 6. EH = 10 CD = 3 BC = 4 AB = 5 1 7 2 7 3 7 4 Exercises 5.6

1. In preparing a certain recipe, a chef uses 5 oz of ingredient

A, 4 oz of ingredient B, and 6 oz of ingredient C. If 90 oz of this dish are needed, how many ounces of each ingredient should be used?

2. In a chemical mixture, 2 g of chemical A are used for each

gram of chemical B, and 3 g of chemical C are needed for each gram of B. If 72 g of the mixture are prepared, what amount in grams of each chemical is needed?

3. Given that , are the following proportions

true? a b AB EF = BD FH AC EG = CD GH AB EF = BC FG = CD GH 4. Given that , are the following proportions true? a b TR XR = SR YR TX XR = RY YS Í XY 7 TS E F G A B C H D S R X T Y

6. Given: , , , ,

Find: FG , GH, EH EF = 6 CD = 4 BC = 5 AB = 7 1 7 2 7 3 7 4 A F B E C D 1 2 3 Exercises 7, 8 C B D A E Exercises 9–12 Exercises 18 and 19 are based on a theorem not stated that is the converse of Theorem 5.6.3.

18. Given: , , ,

and ; and Find: HINT:

19. Given: , , ,

and ; and Find: HINT:

20. Given: In ,

bisects and Find: DC and DB AC = 16 AB = 20 ∠BAC AD 䉭ABC NP MN = PQ MQ . m ∠QNM m ∠M = 36° m ∠P = 62° MQ = 6 PQ = 4 MN = 9 NP = 6 NP MN = PQ MQ . m ∠PNQ m ∠M = 27° m ∠P = 63° MQ = 6 PQ = 3 MN = 8 NP = 4

16. Given: bisects , , ,

Find: WT

17. Given: bisects ,

, , Find: NP MN = 12 QP = 8 NP = MQ ∠MNP NQ WV = 9 UV = 12 WU = 9 ∠WUV UT

13. Given: bisects

Do the following equalities hold? a b

14. Given: bisects

Do the following equalities hold? a b

15. Given: bisects , , ,

Find: TV WT = 6 UV = 12 WU = 8 ∠WUV UT m ∠S = m∠T RS SW = RT WT ∠SRT RW RS RT = SW WT SW = WT ∠SRT RW

10. Given: , , ,

Find: EC

11. Given: , , ,

, Find: a and AD

12. Given: , ,

, , Find: a and EC EC = 3a - 1 AE = a + 1 DB = a + 3 AD = 5 Í DE 7 BC EC = 4a - 5 AE = a DB = 2a + 2 AD = a - 1 Í DE 7 BC AC = 20 DB = 10 AD = 6 Í DE 7 BC

8. Given:

, , , , Find: x , BC, DE

9. Given:

, , , Find: EC AE = 7 DB = 12 AD = 5 Í DE 7 BC EF = 7 DE = x - 2 BC = x AB = 5 1 7 2 7 3

7. Given: , , , ,

Find: x , DE, EF EF = 12 - x DE = x BC = 5 AB = 4 1 7 2 7 3 A F B G E C D H 1 2

3 4

Exercises 5, 6 T S W R Exercises 13, 14 U V T W Exercises 15, 16 N M Q P Exercises 17–19 C A B D

21. In , is trisected

by and so that . Write two different proportions that follow from this information. ∠1 ⬵ ∠2 ⬵ ∠3 CE CD ∠ACB 䉭ABC

33. Use Theorem 5.6.1 and the drawing to complete the proof

of this theorem: “If a line is parallel to one side of a triangle and passes through the midpoint of a second side, then it will pass through the midpoint of the third side.” Given: with M the midpoint of ; Prove: N is the midpoint of RT Í MN 7 ST RS 䉭RST

28. In shown in Exercise 27, suppose that

, and are medians. Find the value of: a b

29. Given point D in the interior of , suppose that

, , , , and . Find RK.

30. Given point D in the interior of , suppose that

, , , and . Find

31. Complete the proof of this property:

If , then and PROOF Statements Reasons 1. 1. ? 2. 2. ? 3. 3. ? 4. 4. ? 5. 5. Means-Extremes Property symmetric form 6. 6. ?

32. Given:

, with , Prove: RX XS = ZT RZ Í YZ 7 RS Í XY 7 RT 䉭RST a + c b + d = c d a + c b + d = a b b a + c = ab + d ab + bc = ab + ad b c = a d a b = c d a + c b + d = c d a + c b + d = a b a b = c d KT KR . HT = 4 SH = 3 GS = 3 RG = 2 䉭RST KT = 3 HT = 5 SH = 4 GS = 4 RG = 3 䉭RST TH HS RK KT SK TG , RH 䉭RST

27. Given point D in the interior of

, which statements is are true? a b TK KR RG GS SH HT = 1 RK KT TH HS GS RG = 1 䉭RST

25. Given: bisects ,