如图:四边形ABCD中,对角线AC、BD相交于点M,△AMB∽△DMC,且AC⊥AB,BD⊥CD,过点A作AE⊥BC,垂足为E,交BD于点F.求证:AB²=BF×BD
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![如图:四边形ABCD中,对角线AC、BD相交于点M,△AMB∽△DMC,且AC⊥AB,BD⊥CD,过点A作AE⊥BC,垂足为E,交BD于点F.求证:AB²=BF×BD](/uploads/image/z/13823441-17-1.jpg?t=%E5%A6%82%E5%9B%BE%EF%BC%9A%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E4%B8%AD%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E3%80%81BD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9M%2C%E2%96%B3AMB%E2%88%BD%E2%96%B3DMC%2C%E4%B8%94AC%E2%8A%A5AB%2CBD%E2%8A%A5CD%2C%E8%BF%87%E7%82%B9A%E4%BD%9CAE%E2%8A%A5BC%2C%E5%9E%82%E8%B6%B3%E4%B8%BAE%2C%E4%BA%A4BD%E4%BA%8E%E7%82%B9F.%E6%B1%82%E8%AF%81%EF%BC%9AAB%26%23178%3B%3DBF%C3%97BD)
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