题目

题型：不详难度：来源：

（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．

答案

证明：
（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
（2）
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