如图,四边形ABCD是矩形纸片,AB=6,对折矩形纸片ABCD,使AD与BC重合,折痕为EF.展平后再过点B折叠矩形纸片,使点A落在EF上的点N,折痕BM与EF相交于点Q,再次展平,连接BN,MN,延长MN交BC于点G.有如下结论:①∠ABN=60°;②AM=3;③△BMG是等边三角形:④EN=3
;⑤P为线段BM上一动点,H是线段BN的动点;则PN+PH的最小值是3
.其中正确结论的序号是 _________ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a7ffe8515ff6183c1c7775dc6f94bdb8.png)
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更新时间:2022-04-03 14:25:58
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