在矩形ABCD中,AB=3,BC=4,点E,F,G,H分别是线段AB,BC,CD,DA上的点,分别以EF,GH所在直线为对称轴,把△BEF,△DGH作轴对称变换得△MEF,△NGH,点M,N恰好在直线AC上,且AM=CN (1)连接BM,MD,DN,NB,求证:四边形BNDM为平
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![在矩形ABCD中,AB=3,BC=4,点E,F,G,H分别是线段AB,BC,CD,DA上的点,分别以EF,GH所在直线为对称轴,把△BEF,△DGH作轴对称变换得△MEF,△NGH,点M,N恰好在直线AC上,且AM=CN (1)连接BM,MD,DN,NB,求证:四边形BNDM为平](/uploads/image/z/2789690-50-0.jpg?t=%E5%9C%A8%E7%9F%A9%E5%BD%A2ABCD%E4%B8%AD%2CAB%3D3%2CBC%3D4%2C%E7%82%B9E%2CF%2CG%2CH%E5%88%86%E5%88%AB%E6%98%AF%E7%BA%BF%E6%AE%B5AB%2CBC%2CCD%2CDA%E4%B8%8A%E7%9A%84%E7%82%B9%2C%E5%88%86%E5%88%AB%E4%BB%A5EF%2CGH%E6%89%80%E5%9C%A8%E7%9B%B4%E7%BA%BF%E4%B8%BA%E5%AF%B9%E7%A7%B0%E8%BD%B4%2C%E6%8A%8A%E2%96%B3BEF%2C%E2%96%B3DGH%E4%BD%9C%E8%BD%B4%E5%AF%B9%E7%A7%B0%E5%8F%98%E6%8D%A2%E5%BE%97%E2%96%B3MEF%2C%E2%96%B3NGH%2C%E7%82%B9M%2CN%E6%81%B0%E5%A5%BD%E5%9C%A8%E7%9B%B4%E7%BA%BFAC%E4%B8%8A%2C%E4%B8%94AM%3DCN+%281%29%E8%BF%9E%E6%8E%A5BM%2CMD%2CDN%2CNB%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E5%9B%9B%E8%BE%B9%E5%BD%A2BNDM%E4%B8%BA%E5%B9%B3)
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