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        1. (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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