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