(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
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