在△ABC中,∠C=90°,AC=CB,AD是∠BAC的平分线,DE⊥AB于E,若AB=10cm,求△BDE的周长

AC=AB*cos∠BAC=50√2
CD=AC*tg∠CAD
=AC*tg(1/2∠BAC)
=AC*(1-cos∠BAC)/sin∠BAC
=5√2*(√2-1)
=10-5√2
BD=BC-CD
=5√2-(10-5√2)
=10(√2-1)
BE=DE
=BD*√2/2
=5(2-√2)
那么周长就是
L=BD+BE+DE
=10（√2-1）+2*5(2-√2)=10.