$ABCD$ is a rectangle of center $O$
$\Rightarrow OA = OB = OC = OD$
Let $H$ be the intersection between $EF$ and $CD$
$ΔAEO$ and $ΔCHO$ have:
$OA = OC$ (proved above)
$\widehat{AOE} = \widehat{COH}$ (vertical angles)
$\widehat{OAE} = \widehat{OCH}$ (anternate interior angles)
Therefore: $ΔAEO=ΔCHO$
$\Rightarrow OE = OH \qquad (1)$
Let $K$ be the midpoint of $BC$
$\Rightarrow BK = KC$
$ΔBCD$ has:
$BK = KC;\, OB = OD$
$\Rightarrow OK$ is the midline
$\Rightarrow OK//CD;\, OK = \dfrac12CD$
$ΔACD$ has:
$AN = ND; \, OA =OC$
$\Rightarrow ON$ is the midline
$\Rightarrow ON//CD;\ ON = \dfrac12CD$
Hence
$O, N, K$ in a line (parallel postulate)
$ON = OK = \dfrac12CD\qquad (2)$
$(1)(2) \Rightarrow ENHK$ is a parallelogram
$\Rightarrow EN//KH \qquad (4)$
$ΔBCD$ has:
$BK = KC;\, CM = MD$
$\Rightarrow KM$ is the midline
$\Rightarrow KM//BD$
Beside: $EF \perp BD$ (assumption)
or $FH\perp BD$
So $FH\perp KM$
Besides: $MH\perp KF \quad (CD\perp BC)$
Thence: $H$ is the orthocenter of $ΔKMF$
$\Rightarrow KH\perp FM\qquad (5)$
$(4)(5)\Rightarrow FM\perp EN$ (Q.E.D)
