如图,已知正方形ABCD,点M是边BA延长线上的动点(不与点A重合),且AM<AB,△CBE由△DAM平移得到.若过点E作EH⊥AC,H为垂足,则有以下结论:①点M位置变化,使得∠DHC=60°时,2BE=DM;②无论点M运动到何处,都有DM=
HM;③无论点M运动到何处,∠CHM一定大于135°.其中正确结论的序号为_____ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cf298f00799cbf34b4db26f5f63af92f.png)
![](https://img.xkw.com/dksih/QBM/2018/7/24/1995294995398656/1996888596676608/STEM/6ccd029082be4f5fbb46303fa1c6e24c.png?resizew=168)
更新时间:2018-07-26 17:03:09
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【推荐1】如图,点E是正方形
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【推荐2】如图,折叠矩形纸片
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【推荐1】如图,正方形
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