菱形ABCD中,AB=4,∠B=60°,E,F分别是AB,AD上的动点,且BE=AF,连接EF,交AC于G,则下列结论:①△BEC≌△AFC;②△ECF为等边三角形;③EF的最小值为2
;④若BE=1,则
=
.其中正确的结论是( )
![](https://img.xkw.com/dksih/QBM/editorImg/2022/6/12/f345ae51-0954-4281-9ee2-83107cd83343.png?resizew=177)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a7ffe8515ff6183c1c7775dc6f94bdb8.png)
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![](https://img.xkw.com/dksih/QBM/editorImg/2022/6/12/f345ae51-0954-4281-9ee2-83107cd83343.png?resizew=177)
A.①② | B.①②③ | C.①②④ | D.①②③④ |
21-22九年级下·湖北咸宁·阶段练习 查看更多[2]
更新时间:2022-06-07 15:44:54
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