矩形ABCD中对角线AC,BD交于点O,AE垂直BD于E,若OE:ED=1:3,AE=根号3则:BD=
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![矩形ABCD中对角线AC,BD交于点O,AE垂直BD于E,若OE:ED=1:3,AE=根号3则:BD=](/uploads/image/z/2395036-28-6.jpg?t=%E7%9F%A9%E5%BD%A2ABCD%E4%B8%AD%E5%AF%B9%E8%A7%92%E7%BA%BFAC%2CBD%E4%BA%A4%E4%BA%8E%E7%82%B9O%2CAE%E5%9E%82%E7%9B%B4BD%E4%BA%8EE%2C%E8%8B%A5OE%3AED%3D1%3A3%2CAE%3D%E6%A0%B9%E5%8F%B73%E5%88%99%EF%BC%9ABD%3D)
矩形ABCD中对角线AC,BD交于点O,AE垂直BD于E,若OE:ED=1:3,AE=根号3则:BD=
矩形ABCD中对角线AC,BD交于点O,AE垂直BD于E,若OE:ED=1:3,AE=根号3则:BD=
矩形ABCD中对角线AC,BD交于点O,AE垂直BD于E,若OE:ED=1:3,AE=根号3则:BD=
设OE=x,则DE=3x,BD=8x,在Rt△ADE 中,由勾股定理得AD=√(9x²+3),
在Rt△ABD中,由勾股定理得AB=√(55x³-3),
∵△ADE∽△BDA,∴3x∶√(9x²+3)=√3∶√(55x²-3),
解得x1=√5/5,X2=-√5/5(舍去),X²=-1/11(舍去).
∴X=√5/5.,∴BD=8√5/5.
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