Given: Right with TEST 534

In Figure 5.44, and are divided proportionally at points B and E if Of course, a pair of segments may be divided proportionally by several points, as shown in Figure 5.45. In this case, and are divided proportionally when notice that 6 4 = 12 8 = 15 10 = 9 6 b RS HJ = ST JK = TV KL = VW LM HM RW AB DE = BC EF or AB BC = DE EF DF AC W V T S R H J K L M 6 12 15 9 4 8 10 6 Figure 5.45 C B D A E Figure 5.46 W V T S R H J K L M Figure 5.47 EXAMPLE 1 In Figure 5.46, points D and E divide and proportionally. If , , and , find AE . Solution , so , where . Then , so . x = AE = 24 7 = 3 3 7 7x = 24 x = AE 4 x = 7 6 AD AE = DB EC EC = 6 DB = 7 AD = 4 AC AB 쮿 A property that will be proved in Exercise 31 of this section is If a b = c d , then a + c b + d = a b = c d In words, we may restate this property as follows: The fraction whose numerator and denominator are determined, respectively, by adding numerators and denominators of equal fractions is equal to each of those equal fractions. Here is a numerical example of this claim: If In Example 2, the preceding property is necessary as a reason. 2 3 = 4 6 , then 2 + 4 3 + 6 = 2 3 = 4 6 EXAMPLE 2 GIVEN: and are divided proportionally at the points shown in Figure 5.47. PROVE: PROOF: and are divided proportionally so that Using the property that if , then , we have Because , , , and , RT HK = TW KM KL + LM = KM TV + VW = TW HJ + JK = HK RS + ST = RT RS HJ = RS + ST HJ + JK = TV + VW KL + LM = TV KL a + c b + d = a b = c d a b = c d RS HJ = ST JK = TV KL = VW LM HM RW RT HK = TW KM HM RW 쮿 Exs. 1, 2 a D E F A B C Figure 5.44 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