如图,在Rt三角形ABC中,角C=90°,BC=2AC,AD是∠BAC的平分线,求证：AB+2BD=5AC

做DF垂直于AB于F点
AB+2BD
=AF+FB+2BD
=AC+2BD+FB
因为 BFD与BCA相似,所以 2FD=FB
所以
AC+2BD+FB
=AC+2BD+2FD
=AC+2CB
=AC+4AC
=5AC

　　延长AC至E，使AE=AB，∵∠BAD=∠EAD，AD=AD，AB=AE，∴△BAD≌△EAD，

　　∴∠E=∠B，又∠ACD=∠ECD，∴△ECD∽△BAD，CE/CD=BC/AC=2

　　∴AB+2BD=AE+2BD=AC+CE+2BD=AC+2CD+2BD=AC+2(CD+BD)=AC+4AC=5AC

过D作DE∥AC，则∠DAC=∠ADE，DA=EA，BE：EA=BD：DC=AB：AC

∴BE=AB*AB/(AB+AC)=5AC^2/(AB+AC) EA=AB*AC/(AB+AC)

∴AB+2BD=BE+EA+2BD=BE+EA+4DE=BE+5EA

=5AC*AC/(AB+AC)+5AC*AB/(AB+AC)=5AC