如图,△ABC中,∠C=90°,AC=BC,AD平分∠CAB交BC于点D,DE⊥AB,垂足为E,且AB=6cm,则△DEB的周长为( )
![](http://hiphotos.baidu.com/zhidao/pic/item/5d6034a85edf8db13e9eefd20a23dd54574e74d0.jpg)
A. 4cm
B. 6cm
C. 8cm
D. 10cm
∵AD平分∠CAB交BC于点D∴∠CAD=∠EAD∵DE⊥AB∴∠AED=∠C=90∵AD=AD∴△ACD≌△AED.(AAS)∴AC=AE,CD=DE∵∠C=90°,AC=BC∴∠B=45°∴DE=BE∵AC=BC,AB=6cm,∴2BC2=AB2,即BC=AB22=622=32,∴BE=AB-AE=AB-AC=6...