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

 

 （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

 

 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