（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．
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AD
CE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
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小学同步三练核心密卷系列答案科目：初中数学 来源： 题型：
如图，在△ABC中，∠ACB=90°，AC=BC=1，取斜边的中点，向斜边作垂线，画出一个新的等腰三角形，如此继续下去，直到所画出的直角三角形的斜边与△ABC的BC重叠，这时这个三角形的斜边为A、
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B、（
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