如上
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∠ADB = ∠AEB 得 ∠ADC = ∠BEC ,∠ACD = ∠BCE 所以△ADC ∽ △BEC 所以得证 CE/BC = CD/ACCE/BC = CD/AC = 2:4 = 1/2S△CDE = ( CE*CD*sin∠C ) / 2S△ABC = ( AC*BC*sin∠C ) / 2 ( CE*CD ) / ( AC*AB ) = 12 : 22 = 1/4 即S△CDE : S△ABC = 1 : 4S△CDE : S□ABDE = S△CDE : ( S△ABC - S△CDE ) = 1 : ( 4 - 1 ) = 1 : 3 = 1 / 3