证明:连接$AC,$交$BD$于点$O。$
$\because$ 四边形$ABCD$是平行四边形,
$\therefore AO=OC,$$BO=DO。$
$\because AM// CN,$$\therefore ∠ EAC=∠ FCA。$
在$△ AEO$和$△ CFO$中,
$\begin{cases}∠ EAO=∠ FCO, \\AO=CO, \\∠ AOE=∠ COF,\end{cases}$
$\therefore △ AEO≌△ CFO。$
$\therefore OE=OF。$
$\therefore BO-OE=OD-OF。$
$\therefore BE=DF$