已知:如图,在Rt△ABC中,∠ACB=90°,AC=BC,点D是BC的中点,CE⊥AD,垂足为点E,BF//AC交CE的延长线于点F.求证:AC=2BF.
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![已知:如图,在Rt△ABC中,∠ACB=90°,AC=BC,点D是BC的中点,CE⊥AD,垂足为点E,BF//AC交CE的延长线于点F.求证:AC=2BF.](/uploads/image/z/4983628-4-8.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8Rt%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0ACB%3D90%C2%B0%2CAC%3DBC%2C%E7%82%B9D%E6%98%AFBC%E7%9A%84%E4%B8%AD%E7%82%B9%2CCE%E2%8A%A5AD%2C%E5%9E%82%E8%B6%B3%E4%B8%BA%E7%82%B9E%2CBF%2F%2FAC%E4%BA%A4CE%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9F.%E6%B1%82%E8%AF%81%EF%BC%9AAC%3D2BF.)
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