设AC,DE交点为F,过D作DG垂直AC于G点由题意易知:∠DAE=∠ADE=45,则AD=6,DE=3√2,也可求得DE,EF,DG=AD/(1+√2),AF容易证:三角形CDE相似于三角形FDC,则CD2=DE*DF,CG2=CD2-DG2,GF=DF/(1+√2),CF=CG+GF,又由BE/AE=CF/AF,可求得BE