如图,在矩形ABCD中,E是AD的中点,BE⊥AC于点F,连接DF,分析下列四个结论:①△AEF∽△CAB;②CF=2AF;③DF=CD;④S四边形CDEF=
S△ABF.其中正确的是( )
![](https://img.xkw.com/dksih/QBM/2022/5/16/2980516332118016/2980736493043712/STEM/08baaab1-8d09-4bc3-96ec-90b9b63038b7.png?resizew=167)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/533a7b702ada1dd80123e4041271d521.png)
![](https://img.xkw.com/dksih/QBM/2022/5/16/2980516332118016/2980736493043712/STEM/08baaab1-8d09-4bc3-96ec-90b9b63038b7.png?resizew=167)
A.①② | B.①②③ |
C.①②③④ | D.① |
更新时间:2022-05-16 17:46:49
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