做EH∥AB交CB于H∴△FEH∽△FDB∴EF/DF=EH/BD∵EH∥AB∴△CEH∽△CAB∴CE/AC=EH/AB。即EH/CE=AB/AC∵CE=BD，那么EH/BD=EH/CE∴EF/DF=AB/AC即AB×DF=AC×EF
过点E作EG平行于AC。EG/AB=CE/CA，EG/BD=EF/DF，BD=CE可以推出，AB*DF=AC*EF
考虑一下
