题目

（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 

查看本题答案和解析