在△ABC中,AE是BC上的中线,F为AE上任意一点,连结CF并延长交AB于D,试说明AD/AB=DF/FC

过D作DN//BC,交AE于M,
〈DAN=〈BAC,
〈ADN=〈ABC,（同位角相等）
△ADN∽△ABC,
AD/AB=DN/BC,
同理,DM/BE=AM/AE,
MN/EC=AM/AE,
DM/BE=MN/EC,
而BE=CE,
故DM=MN,
故AD/AB=DM/CE,
〈MDF=〈ECF,（内错角相等）,
〈DFM=〈CFE,（对顶角相等）,
△DMF∽△CEF,
故DF/CF=DM/CE,
故AD/AB=DF/CF.