如图,等腰直角三角形ABC中,CA=CB,点E为△ABC外一点,CE=CA,且CD平分∠ACB交AE于D,∠CDE=60°.求证:△CBE为等边三角形.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/7/21/7d313777-af78-4d9e-a8a8-0a301e139568.png?resizew=172)
更新时间:2018-08-27 21:51:54
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