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在直角梯形ABCD中,AD∥BC,∠ABC=90°,AB=BC,E为AB边上一点...
在直角梯形ABCD中,AD∥BC,∠ABC=90°,AB=BC,E为AB边上一点,∠BCE=15°,且AE=AD.连接DE交对角线AC于H,连接BH.下列结论:
①△ACD≌△ACE;②△CDE为等边三角形;③
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131101185840494701770/SYS201311011858404947017009_ST/0.png)
=2;④
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131101185840494701770/SYS201311011858404947017009_ST/1.png)
.
其中结论正确的是( )
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131101185840494701770/SYS201311011858404947017009_ST/images2.png)
A.只有①②
B.只有①②④
C.只有③④
D.①②③④
考点分析:
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