$ABCD$ 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)