推 littleme1125: 感謝您! 10/01 18:52
※ 引述《littleme1125 (..)》之銘言:
: 三角形ABC中,角A為直角,BC中點為D,圓O過A,C,D三點
: AE為三角形ABC的高,直線BO交直線AE於點F,則 線段EF/線段AF =?
: 我用ABC為等腰直角下去算,可以求出答案為 1/2
: 有比較正規的做法嗎?
國中作法
OD交AE於P,AC中點H
AC = b, AB = c, BC = a
BE = (1/a)c^2
DE = a/2 - (1/a)c^2 = [a^2 - 2c^2]/(2a) = [b^2 - c^2]/(2a)
DP = [b^2 - c^2]/(2c)
EP = (b/c)[b^2 - c^2]/(2a)
OD = (a/c)(a/4) = (a^2)/(4c)
OP : AB = (EP + EF) : FA
=> (3b^2 - c^2)/(4c) : c = {[(b^3 - bc^2)/(2ac)] + EF]} : FA
=> [3b^2 - c^2]FA = (2bc/a)(b^2 - c^2) + 4c^2 EF -------(1)
DP : AB = EP : (EF + FA)
=> (b^2 - c^2)/(2c) : c = [b^3 - bc^2]/(2ac) : [EF + FA]
=> [b^2 - c^2][EF + FA] = (bc/a)(b^2 - c^2) ---------(2)
由(1), (2)
(b^2 + c^2)FA = 2(b^2 + c^2)EF
=> EF/FA = 1/2
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