This article provides three methods of proof.
Proof 1: As shown in Figure 2, draw $PA$ and $QD$ perpendicular to $AD$, intersecting $EF$ at $P$ and $Q$ respectively.
On $BC$, mark $BJ$
$$= AP, KC = DQ \text{, }$$

Connect $AJ$ and $DK$, then
$DQ \parallel MN \parallel PA$.
Since $AN = ND$, it follows that $PM = MQ$.
Also, $\angle EAP + \angle BAD = \angle BAD + \angle ABC = 180^{\circ}$, thus
$\angle EAP = \angle ABJ$.
In $\triangle AEP$ and $\triangle BAJ$,
$\angle EAP = \angle ABJ, AE = AB, AP = BJ$,
Therefore, $\triangle AEP \cong \triangle BAJ$.
Hence $EP = AJ, \angle EPA = \angle BJA$.
Thus, $\angle APQ = \angle AJK$.
Similarly, $FQ = DK, \angle FQD = \angle DKC$.
Since $PA \parallel QD$, then
$$\angle FQD = \angle QPA = \angle AJK = \angle DKC .$$

Therefore, $DK \parallel AJ$.
Thus, quadrilateral $AJKD$ is a parallelogram.
Hence $AJ = KD = EP = QF$.
Therefore, $EP + PM = MQ + QF$, i.e.,
$ME = MF$.
Thus, $M$ is the midpoint of $EF$.
Proof 2: As shown in Figure 3, draw $AP$ and $DQ$ perpendicular to $AD$, intersecting $BC$ at $P$ and $Q$ respectively.
Draw perpendiculars from $E$ and $F$ to $AD$, intersecting the line containing $AD$ at $K$ and $N$ respectively.
Since $AP \perp AD, \angle KAB + \angle BAP = 90^{\circ}$, hence $\angle EAK = \angle BAP$.
Since $AE = AB, \angle AKE = \angle APB$, therefore,
$\triangle AKE \cong \triangle APB$.
Thus, $AP = AK$.
Similarly, $DQ = DN$.
Since $AD \parallel BC$, therefore,
$AP = DQ = AK = DN$.
Since $AJ = DJ$, therefore,
$JK = JN$.
Since $EK \parallel MJ \parallel FN$, therefore,
$EM = MF$.
Hence $M$ is the midpoint of $EF$.
Proof 3: As shown in Figure 4, draw $ET \perp l$ at $T$, $FR \perp l$ at $R$, draw $IP \perp ET$ intersecting $BC$ at $P$, and draw $SQ \perp RF$ intersecting $BC$ at $Q$.

It is easy to see that quadrilaterals TIAN, TIPK, RSDN, and RSQK are all rectangles.
Therefore, $IT = AN = ND = RS$,
$$\begin{array}{l} \angle EAI = 90^{\circ} - \angle BAP = \angle ABP, \\ \angle EIA = \angle APB = 90^{\circ}, AE = AB . \end{array}$$

Therefore, $\triangle EIA \cong \triangle APB$. Thus, $EI = AP$.
Similarly, $FS = DQ$.
It is also easy to see that quadrilateral $APQD$ is a rectangle, so $AP = DQ$.
Therefore, $EI = AP = DQ = FS$.
Thus, $EI + IT = FS + SR$, i.e., $ET = FR$.
Since $ET \parallel FR$, therefore, $EM = FM$.
Thus, $M$ is the midpoint of $EF$.