※ 引述《mj813 (薩坨十二惡皆空)》之銘言:
: 菱形ABCD,角B為60度。
: 直線EF通過D點,
: 分別與BA、BC的延長線交於E、F。
: 設M為CE與AF的交點,
: 若CM=4,EM=5,求AC=?
: 麻煩各位前輩解惑,感激不盡!
設AF交CD於N
AC = a
AE = t
CF = k
△EAD ~ △DCF => a^2 = kt
CN = ak/(a + k)
令x = EM/MC = 5/4
=> x = t/[ak/(a + k)] = a(a + k)/(k^2)
=> a/k = [-1 + √(1 + 4x)]/2 = [-1 + √6]/2
=> t = (1/k)a^2 = a[-1 + √6]/2
=> a^2 + (1/4)a^2[1 + √6]^2 - a^2 [-1 + √6]/2 = 81
=> AC = a = 6
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 111.243.49.54
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1511791894.A.5CE.html