Question

（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．

Answer

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