等腰梯形ABCD中,AD平行BC,AB=DC,AC⊥BD,过D点作DE平行AC交BC的延长线于E点.问：若AD＝3,BC
3个回答

因为,AD∥CE,DE∥AC,

所以,ACED是平行四边形,

可得：CE = AD = 3 ,S△ACD = (1/2)S平行四边形ACED = S△CDE ,DE = AC .

因为,△ABD和△ACD同底等高,

所以,S△ABD = S△ACD = S△CDE ,

可得：S梯形ABCD = S△ABD+S△BCD = S△CDE+S△BCD = S△BDE .

过点D作DF⊥BC于F.

因为,AC⊥BD,DE∥AC,

所以,DE⊥BD,

而且,BD = AC = DE ,

则有：△BDE为等腰直角三角形,

可得：斜边 BE = BC+CE = 10 ,斜边上的高 DF = (1/2)BE = 5 ,

所以,S梯形ABCD = S△BDE = (1/2)·BE·DF = 25 .

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