請問這題等腰梯形角度求解?(OMC複試)
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Re: 請問這題等腰梯形角度求解?(OMC複試)
ABCD 是圓內接四邊形,AC 是 ∠BAD 的平分線,CD = BC = (1/2)AD
令 △AFD = x
△ACD 和 △AED 同底(AD),面積比 = 高的比
△ACD/△AED = [(√3/2)CD]/[(√3/2)AE] = CD/AE
(x + 70)/(x + 16) = CD/AE
△AFD/△AFE = FD/FE = AD/AE (內分比定理)
x/16 = 2CD/AE
(x/2)/16 = CD/AE
(x + 70)/(x + 16) = (x/2)/16
x = 56
△AFD = 56
令 △AFD = x
△ACD 和 △AED 同底(AD),面積比 = 高的比
△ACD/△AED = [(√3/2)CD]/[(√3/2)AE] = CD/AE
(x + 70)/(x + 16) = CD/AE
△AFD/△AFE = FD/FE = AD/AE (內分比定理)
x/16 = 2CD/AE
(x/2)/16 = CD/AE
(x + 70)/(x + 16) = (x/2)/16
x = 56
△AFD = 56