如图,已知三角形ABC,延长BC到D,使CD=BC,取AB中点F,连接FD交AC于点E.求AE比AC的值

过点F作FG‖BC,交AC于点G.
因为,FG‖BC,AF = FB = (1/2)AB ,
所以,AG = GC = (1/2)AC ,FG = (1/2)BC .
因为,FG‖CD,FG = (1/2)BC = (1/2)CD ,
所以,GE = (1/2)EC .
因为,GE∶EC = 1∶2 ,
可得：GE∶GC = GE∶(GE+EC) = 1∶(1+2) = 1∶3 ,
所以,GE = (1/3)GC = (1/6)AC ,
可得：AE = AG+GE = (2/3)AC ,
即有：AE∶AC = 2/3 .