如图,在△ABC中,∠A＝90°,AB=AC,BE平分∠ABC交AC于D,CE⊥BD的延长线于E,求证BD=2CE.

因BD平分∠ABC ,故AD/CD=AB/BC=1/√2 ,若AD=1,则CD=√2 ,即AB=1+√2 ,故BD^2= AB^2+AD^2=(1+√2)^2+1=2( 2+√2) ,BD=√2( 2+√2) 又CD⊥BD,AB⊥AC ,故∠ABD=∠ECD ,故Rt△ABD∽Rt△ECD ,CE=CD*AB/BD=√2 (1+√2 )/√2( 2+√2) = (1+√2) / √( 2+√2) ,故BD/CE=√2( 2+√2) / [(1+√2) / √( 2+√2) ] = (2+2√2)/(1+√2) =2 ,即BD=2CE.