小學數一題~~20點

△ABE ~ △CFE
因AE : EC = 3 : 2
得AB : FC = 3 : 2(對應邊成比例)......(1)
又 已知 AB : DC = 2 : 3 ......(2)
由 (1) , (2) 得
AB = (3/2)FC , 及 AB = (2/3)DC
(3/2)FC = (2/3)DC......(= AB)
FC : DC = 4 : 9
△AEF = △ACF * AE/AC = △ACF * AE/(AE+EC)

= △ACF * 3/(3+2) = △ACF * 3/5
= (△ACD * FC/DC) * 3/5
= (△ACD * 4/9) * 3/5
= △ACD * 4/15
= [全梯形 * DC/(AB + DC)] * 4/15
= [100 * 3/(2+3)] * 4/15
= 16 cm^2

我看到喔.....................

2010-04-10 17:01:00 補充：
你還是用樓下的解答吧~
我的答案可能是錯的

死圖...