梯形ABCD中,AB‖DC,过点D作DE‖CB,交CA延长线于点E,BD与AC相交于点O.求证:OC²=OA·
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梯形ABCD中,AB‖DC,过点D作DE‖CB,交CA延长线于点E,BD与AC相交于点O.求证:OC²=OA·OE.
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![梯形ABCD中,AB‖DC,过点D作DE‖CB,交CA延长线于点E,BD与AC相交于点O.求证:OC²=OA·](/uploads/image/z/19086034-58-4.jpg?t=%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAB%E2%80%96DC%2C%E8%BF%87%E7%82%B9D%E4%BD%9CDE%E2%80%96CB%2C%E4%BA%A4CA%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9E%2CBD%E4%B8%8EAC%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9O.%E6%B1%82%E8%AF%81%EF%BC%9AOC%26%23178%3B%3DOA%C2%B7)
AB‖DC,则⊿ABO∽⊿CDO,得:OC/OD=OA/OB ----(1)
DE‖CB,则⊿BCO∽⊿DEO,得:OC/OB=OE/OD ----(2)
(1)×(2)得:OC²=OA·OE.
再问: 有木有详细点儿的
DE‖CB,则⊿BCO∽⊿DEO,得:OC/OB=OE/OD ----(2)
(1)×(2)得:OC²=OA·OE.
再问: 有木有详细点儿的
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