题目内容
（6分）如图，在△ABC中，AB=AC，AD⊥BC，垂足为D，AE∥BC, DE∥AB.
证明：（1）AE=DC；（2）四边形ADCE为矩形．
（1）在△ABC中，∵AB=AC，AD⊥BC，
∴BD=DC························································································································ 1分
∵AE∥BC, DE∥AB，
∴四边形ABDE为平行四边形························································································· 2分
∴BD=AE，····················································································································· 3分
∵BD=DC
∴AE = DC．··················································································································· 4分
解法一：∵AE∥BC，AE = DC，
∴四边形ADCE为平行四边形．······················································································ 5分
又∵AD⊥BC，
∴∠ADC=90°，
∴四边形ADCE为矩形．································································································ 6分
解法二：
∵AE∥BC，AE = DC，
∴四边形ADCE为平行四边形························································································· 5分
又∵四边形ABDE为平行四边形
∴AB=DE．∵AB=AC，∴DE=AC．
∴四边形ADCE为矩形．································································································ 6分
解析:略
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