（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．
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
科目：初中數學 來源： 題型：
如圖，在△ABC中，∠ACB=90°，AC=BC=1，取斜邊的中點，向斜邊作垂線，畫出一個新的等腰三角形，如此繼續下去，直到所畫出的直角三角形的斜邊與△ABC的BC重疊，這時這個三角形的斜邊為A、
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