Given: with angles TEST 534

33. Given: with , ,

and Find: AB HINT: Use altitude from C to as an auxiliary line. AB CD BC = 12 m ∠B = 30° m ∠A = 45° 䉭ABC A B C 30° 45° M N T Q P 12 120°

34. Given:

Isosceles trapezoid MNPQ with and ; the bisectors of MQP and NPQ meet at point T on Find: The perimeter of MNPQ MN ∠s m ∠M = 120° QP = 12

35. In regular hexagon ABCDEF,

. Find the exact length of: a Diagonal b Diagonal

36. In regular hexagon ABCDEF, the

length of AB is x centimeters. In terms of x, find the length of: a Diagonal b Diagonal CF BF CF BF AB = 6 inches F E B A D C Exercises 35, 36 Y V Z X W D E C A B

37. In right triangle XYZ, and . Where

V is the midpoint of and , find VW. HINT: Draw XV. m ∠VWZ = 90° YZ YZ = 4 XY = 3

38. Diagonal separates pentagon

ABCDE into square ABCE and isosceles triangle DEC. If and , find the length of diagonal . HINT: Draw DF ⬜ AB. DB DC = 5 AB = 8 EC Segments Divided Proportionally The Angle-Bisector Theorem Ceva’s Theorem Segments Divided Proportionally 5.6 KEY CONCEPTS In this section, we begin with an informal description of the phrase divided proportion- ally . Suppose that three children have been provided with a joint savings account by their parents. Equal monthly deposits have been made to the account for each child since birth. If the ages of the children are 2, 4, and 6 assume exactness of ages for sim- plicity and the total in the account is 7200, then the amount that each child should re- ceive can be found by solving the equation Solving this equation leads to the solution 1200 for the 2-year-old, 2400 for the 4-year- old, and 3600 for the 6-year-old. We say that the amount has been divided proportion- ally. Expressed as a proportion, this is 1200 2 = 2400 4 = 3600 6 2x + 4x + 6x = 7200 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