过A点作BC的垂线AG,交BC于G,作CD的垂线AH,交CD于H.
S△ABC=(1/2)*AG*BC;
S△ABE=(1/2)*AG*BE;
S△ADC=(1/2)*AH*CD;
S△AFD=(1/2)*AH*FD.
所以S△ABC/S△ABE=〔(1/2)*AG*BC〕/〔(1/2)*AG*BE〕=BC/BE;
同理,△ADC/S△AFD=CD/FD.
又由EF//BD,得出BC/BE=CD/FD;
即:S△ABC/S△ABE=△ADC/S△AFD;
由平行四边形可知S△ABC=S△ADC;
所以S△ADF=S△ABE