Question

（6分）如圖，在△ABC中，AB=AC， AD⊥BC，垂足為D，AE∥BC, DE∥AB.

 

證明：（1）AE=DC；

（2）四邊形ADCE為矩形．

 

Answer 1

證明：

（1）在△ABC中，∵AB=AC，AD⊥BC，

∴BD=DC······························································································ 1分

∵AE∥BC, DE∥AB，

∴四邊形ABDE為平行四邊形······································································ 2分

∴BD=AE，···························································································· 3分

∵BD=DC

∴AE = DC．·························································································· 4分

（2）

解法一：∵AE∥BC，AE= DC，

∴四邊形ADCE為平行四邊形．··································································· 5分

又∵AD⊥BC，

∴∠ADC=90°，

∴四邊形ADCE為矩形．··········································································· 6分

解法二：

∵AE∥BC，AE= DC，

∴四邊形ADCE為平行四邊形······································································ 5分

又∵四邊形ABDE為平行四邊形

∴AB=DE．∵AB=AC，∴DE=AC．

∴四邊形ADCE為矩形．··········································································· 6分

解析:略