已知梯形ABCD中,AB‖CD,AD=BC=DC=2,cosA=3/4(四分之三),M为边BC上的一点,且AM=1.(1)求AB的长.(2)求证:CM=2DM(3)将∠DMC绕点M顺时针旋转后,得到∠D1MC1(点D1、C1依次与点D、C对应),射线MD1交线
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![已知梯形ABCD中,AB‖CD,AD=BC=DC=2,cosA=3/4(四分之三),M为边BC上的一点,且AM=1.(1)求AB的长.(2)求证:CM=2DM(3)将∠DMC绕点M顺时针旋转后,得到∠D1MC1(点D1、C1依次与点D、C对应),射线MD1交线](/uploads/image/z/4105869-69-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAB%E2%80%96CD%2CAD%3DBC%3DDC%3D2%2CcosA%3D3%2F4%EF%BC%88%E5%9B%9B%E5%88%86%E4%B9%8B%E4%B8%89%EF%BC%89%2CM%E4%B8%BA%E8%BE%B9BC%E4%B8%8A%E7%9A%84%E4%B8%80%E7%82%B9%2C%E4%B8%94AM%3D1.%EF%BC%881%EF%BC%89%E6%B1%82AB%E7%9A%84%E9%95%BF.%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9ACM%3D2DM%EF%BC%883%EF%BC%89%E5%B0%86%E2%88%A0DMC%E7%BB%95%E7%82%B9M%E9%A1%BA%E6%97%B6%E9%92%88%E6%97%8B%E8%BD%AC%E5%90%8E%2C%E5%BE%97%E5%88%B0%E2%88%A0D1MC1%EF%BC%88%E7%82%B9D1%E3%80%81C1%E4%BE%9D%E6%AC%A1%E4%B8%8E%E7%82%B9D%E3%80%81C%E5%AF%B9%E5%BA%94%EF%BC%89%2C%E5%B0%84%E7%BA%BFMD1%E4%BA%A4%E7%BA%BF)
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