xiaoqianho123
幼苗
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延长DE,和CB的延长线相交于点F.
已知,AD∥BC,可得:∠ADE = ∠BFE ;
因为,在△ADE和△BFE中,∠ADE = ∠BFE ,∠AED = ∠BEF ,AE = BE ,
所以,△ADE ≌ △BFE ,
可得:DE = EF ,S△ADE = S△BFE ;
S△CDE = S△CEF = (1/2)S△CDF ,
S四边形ABCD = S△ADE+S四边形BCDE = S△BFE+S四边形BCDE = S△CDF = 2S△CDE .
1年前
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