Beautiful Problem! WLOG, assume $AB<CA$. Let $H$ be the orthocenter WRT $\Delta ABC$. Let $H'$ be the reflection of $H$ over $BC$ and $F$ be the midpoint of $DH'$ $\implies$ $FE$ $||$ $H'Q$. Let $G$ be midpoint of $BC$ and if $O$ is the center of $\odot (ABC)$ $\implies$ $OE$ $||$ $AD$ $\implies$ $E$ lies on perpendicular bisector of $BC$ $\implies$ $DFEG$ is a rectangle. Also, since, $NG=MB=MD$ and $\angle MDF$ $=$ $90^{\circ}$ $+$ $\angle ABC$ $=$ $90^{\circ}$ $+$ $\angle NGC$ $=$ $\angle NGE$ $\implies$ $\Delta MDF$ $\cong$ $\Delta NGE$ $\implies$ $MF=NE$ and $MN||FE$ $\implies$ $MFEN$ is cyclic. Let $\odot (MFEN)$ $\cap$ $\overline{AD}$ $=$ $X$ $\implies$ $\angle AFE$ $=$ $90^{\circ}$ $=$ $\angle XME$ $=$ $\angle XNE$. Hence, $X$ is the desired point on $AD$

Solution. Basically, the same solution of #2. Let $D'=\overline{AD}\cap \Gamma,\ D'\neq A$, $F$ the midpoint of $DD'$, $L$ the midpoint of $BC$ and $O$, as usual, the circumcenter of $\bigtriangleup ABC$. Clearly, $OE\parallel AD$ so $OE\perp BC$ and thus $O,\ L$ and $E$ are collinear. Moreover, since $\angle DD'Q=90^\circ=\angle ADL$ and $EF\parallel D'Q$ we get $BC\parallel EF\parallel D'Q$. We infer that $DLEF$ is a rectangle. From here, the conclusion is clear: recall that $MNLD$ is an isosceles trapezoid because $MN\parallel DL$ and these four points lie on the nine-point circle of $\bigtriangleup ABC$; therefore, $MN$ and $DL$ has the same perpendicular bisector $\ell$, which also must bisects $FE$ perpendicularly because of the rectangle $DLEF$; hence, $MNEF$ is an isosceles trapezoid. Thus, if $P$ is the second intersection of $AF$ and $(MNEF)$ we must have $$\angle ENP=\angle EMP=\angle EFP=90^\circ$$as required. $\blacksquare$

This problem was proposed by me. The solution I sent is completely ugly compared with the ones I have seen, and doesn't deserve time to be read. Anyway, here it is: Solution. Let $H$ and $O$ be the orthocenter and circumcenter of triangle $ABC$, $I$ the intersection point of $AD$ and $MN$, and $J$ the intersection point of the perpendicular to $BC$ passing through $O$ (which is also perpendicular to $MN$, since $MN\parallel BC$). Observe that $O$ is the midpoint of $AQ$. Then, in triangle $ADQ$, $OE$ is a line parallel to $AD$, so $E$ lies on the line perpendicular to $BC$ passing through $O$. We know that the center of the nine-point circle (passing through $M$ and $N$) is the midpoint of $OH$, and lies on the perpendicular bisector of $MN$, so it bisects $OH$, from which $AD$ and $OE$ are lines reflected over this perpendicuar bisector, and thus $MI=JN$. Since $EJ\perp MN$, we have $EM^2-EN^2=JM^2-JN^2$, and since $(MM^2-EM^2)+(EN^2-NN^2)+(NI^2-MI^2)=(EN^2-EM^2)-(IM^2-IN^2)= (EN^2-EM^2)-(JN^2-JM^2)=0$, by Carnot's Theorem, it follows that the perpendicular from $I$ on $MN$, from $M$ on $ME$ and from $N$ on $NE$ concur, or, seen in another way, the perpendiculars to $EM$ and $EN$ through $M$ and $N$ meet on $AD$.

Let $AD\cap (O)=A,K$ and $P$ be the midpoint of $KD$. We show that $M,N,E,P$ lies on a circle. Then $EP\| BC\| KQ\| MN$ so $MNEP$ is a trapezoid. We have to prove that $ME=NP$. We have $ME^2=\dfrac{2(MD^2+MQ^2)-DQ^2}{4}$ and $NP^2=\dfrac{2(ND^2+NK^2)-DK^2}{4}$ so $ME^2=MP^2\Leftrightarrow 2(MD^2-ND^2+MQ^2-NK^2)+DK^2-DQ^2=0=S$. Note that $KC=BQ$. We have $MQ^2-NK^2=\dfrac{2(AQ^2+BQ^2)-AB^2}{4}-\dfrac{2(AK^2+KC^2)-AC^2}{4} =\dfrac{2AQ^2-2AK^2-AB^2+AC^2}{4}$. And $MD^2-ND^2=\dfrac{AB^2-AC^2}{4}$ so $S=AQ^2-AK^2+DK^2-DQ^2=KQ^2-KQ^2=0$. This means $MNEP$ is a scale trapezoid. Otherwise $EP\perp AD$ then Q.E.D

Nice problem! Take $M'$ and $N'$ to be the reflections of $E$ over $M$ and $N$. The perpendiculars to $EM$ and $EN$ passing through $M$ and $N$ meet in the circumcenter of triangle $EM'N'$, so it is enough to prove that $AD$ is the perpendicular bisector of $M'N'$. First, $MN$ is parallel to $M'N'$ and $AD$ is perpendicular to $MN$, so $AD$ is perpendicular to $M'N'$. Now we show that $AM' = AN'$ which will complete the proof. Notice that if $O$ is the circumcenter of $ABC$, then since $OE$ is parallel to $AD$, $OE$ is perpendicular to $BC$, and therefore $E$ is in the perpendicular bisector of $BC$, so that $BE = CE$. Also, $AM'BE$ and $AN'CE$ are parallelograms, by construction. Hence $AM' = BE = CE = AN'$ and we are done.

Here's a simple solution: Lemma. Let $ABC$ be a triangle and let $A'$ be the $A$-antipode. Then the lines perpendicular to $BC$ through $A$ and $A'$ are symmetric wrt the perpendicular bisector of $BC$. Proof. Let $O$ be the circumcenter of $ABC$, which lies on the perpendicular bisector of $BC$. Then $A$ and $A'$ are symmetric wrt $O$, so the perpendiculars from them to $BC$ are symmetric wrt the perpendicular bisector of $BC$. Now, let $X, Y$ and $Z$ be the midpoints of $DA, DB$ and $DC$ respectively. Then by homothety, $E$ is the $X$-antipode in $(XYZ)$, so by the Lemma the perpendiculars to $YZ$ through $X$ and $E$ are symmetric wrt the perpendicular bisector of $YZ$. By the Lemma again, the $E$-antipode $W$ in $(EMN)$ is such that the lines perpendicular to $MN$ through $E$ and $W$ are symmetric wrt the perpendicular bisector of $MN$. However $MN$ and $YZ$ have the same perpendicular bisector, so by the two previous applications, $WX \perp MN$. As $X$ lies on $AD$ and $AD \perp MN$ it follows that $W$ lies on $AD$, as desired.

Another solution. We'll prove the following more general result. Lemma 1. Let $M,\ N$ and $D$ be defined in the same way as they we're given in the problem statement. Let $P$ be an arbitrary point on the perpendicular bisector of the side $BC$ in a triangle $ABC$ so that $M,\ P$ and $N$ are not collinear. Then, the perpendicular lines through $M$ and $N$ to $MP$ and $NP$, respectively, meet on $AD$. Proof. Let $L$ be the midpoint of $BC$. As argued in post #4, $MNLD$ is an isosceles trapezoid. Since $AD$ and $PL$ are perpendicular to $BC$, these lines are symmetric with respect to the perpendicular bisector of $MN$ and $DL$. Thus, if $Q=\overline{AD}\cap (MPN)$ so that $Q$ and $P$ lie on the same side with respect to $MN$, we must have $QP\parallel MN\parallel BC$, therefore $\angle DQP=90^\circ$. Defining $R$ as $\overline{QD}\cap (MPN),\ R\neq A$, it follows that $\angle RMP=\angle RNP=\angle RQP=90^\circ$, as desired. $\blacksquare$ Clearly, the lemma above implies that our solution is complete, since $E$ lies on the perpendicular bisector of $BC$. We're done! $\blacksquare$

One-liner solution from a countryman: since $\Delta EMN$ and $\Delta ANM$ are orthologic, we're done.

One-liner solution from a refugee: Let \(AD \cap (O) = R, P=\) midpoint of \(DR\) , \(l || AD : N_9 \in l\). \(M,N\) anD \(P, E\) are the reflections in \(l\) so, cyclic and antipode \(E \in AD\), we're done.

When you don't have ideas. Let's do it by coordinate menthod. Suppose $A_{(0,a)}$, $B_{(b,0)}$, $C_{(c,0)}$. The lines perpendicular to $EM$ and $EN$ passing through $M$ and $N$ respectively meets at $P$. Remark 1. $D_{(0,0)}$ Remark 2. $Q_{\left (c+b,\frac{cb}{a} \right )}$. So $E_{\left ( \frac{c+b}{2},\frac{cb}{2a} \right )}$. Remark 3. $PM: (x-b/2).c/2+(y-a/2)(cb/a-a/2)=0.$ $PN: (x-c/2).b/2+(y-a/2)(cb/a-a/2)=0.$ So it is easy to see that point $P$ has abscissa is 0 or $P$ lies on $AD$. Q.E.D. P/S: I type this solution by my phone so i abbreviate latex formula.

let $K$ intersection of $AD$ and $\Gamma$ and $P$ midpoint of $DK$ .let $Y=AD\cap MN$ and let say $R$ on $MN$ such that $\angle ORN=90^\circ$ and let say $T$ on $BC$ such that $\angle QTC=90^\circ$. $\angle QAC=90^\circ-\angle OQC=90^\circ-\angle ABC=\angle BAK$ so $KQ\parallel BC$ and $BK=QC$ so $BKQC$ trapozoid and $BD=CT$. if we apply homotety with center $A$ and ratio $1/2$ points $A,M,N$ goes to $A,B,C$ and $O$ goes to $Q$ so we have $R$ goes to $T$ and $NR=\dfrac{CT}{2}$. Also in this homotety point $Y$ goes to $D$ and we have $MY=\dfrac{BD}{2}=\dfrac{CT}{2}=NR$. $DE=EQ$ and $AO=OQ$ so we have $AD\parallel EO$. Because of $AD\parallel EO$ ,$EO$ perpendicular to $BC$ so $R,E,O$ on same line. because of $M,N$ midpoints, $MN\parallel BC$ and because of $P,E$ midpoints of $DK,DQ$ we have $PE\parallel KQ\parallel BC\parallel MN$. Because of $PE\parallel MN$ we have $PY=ER$.Because of $PE\parallel KQ$ and $AQ$ diameter we have $\angle YPE=\angle DKQ=90^\circ$ . so we have $MY=NR$ , $PY=ER$ and $\angle PYM=\angle ERN=90^\circ$. From there we have $MYP$ and $NRE$ similar and we found $PE\parallel MN$ , so $\angle PMN=\angle MNE=180^\circ-\angle PEN$. From there we have $PENM$ cyclic. let $(PENM)\cap AD=P,Z$ so because of $\angle ZPE=90^\circ$ we have $\angle ZME=90^\circ$ and $\angle ZNE=90^\circ$ so we get result.

Let $S$ such that $SM\perp ME$ and $SN\perp NE $, $(X,Y)=MN\cap (AD,EO)$ and $P\in EO $ such that $PN\parallel ME $. Now let $X'\in MN$ such that $SX'\perp MN $. We will show that $X=X'$. First see that $EO\parallel AD\implies EO\perp MN $ with this: $$NY=\frac {BC}{2}-\frac {DC}{2}=\frac {BD}{2}=MX $$ Now see that $\triangle MSN\sim \triangle PNE $ yields: $$\frac {MX'}{X'N}=\frac {PY}{YE}=\frac {NY}{YM}~\text{(since PN $\parallel$ ME) }$$ This means $\frac {MX'}{X'N}=\frac {MX}{XN} $ since $MX=NY $. This implies $X'=X $.

Jafet98 wrote: Let $ABC$ be a triangle and $\Gamma$ its circumcircle. Let $D$ be the foot of the altitude from $A$ to the side $BC$, $M$ and $N$ the midpoints of $AB$ and $AC$, and $Q$ the point on $\Gamma$ diametrically opposite to $A$. Let $E$ be the midpoint of $DQ$. Show that the lines perpendicular to $EM$ and $EN$ passing through $M$ and $N$ respectively, meet on $AD$. Let $P$ is the intersection of the lines perpendicular to $EM$ and $EN$ passing through $M$ and $N,$ note that $OE \perp MN$ we have \[PM^2-PN^2=EN^2-EM^2=ON^2-OM^2=AM^2-AN^2.\]That mean $AP \perp MN,$ so $P \in AD.$