[asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A = dir(100), B = dir(220), C = dir(320), H = orthocenter(A,B,C), D = foot(A,B,C), E = foot(C,A,B), M = (E+D)/2, N = intersectionpoint(M--M+dir(A--M)*100,unitcircle), Ha = 2D-H, Hb = 2*foot(B,C,A)-H, K = extension(A,A+dir(D--E),Hb,foot(Hb,B,C)); draw(A--B--C--A); draw(unitcircle); draw(A--Ha^^B--Hb^^C--E, gray); draw(D--E^^A--N); draw(N--Hb, dashed); draw(A--K--Hb); dot("$A$", A, dir(100)); dot("$B$", B, dir(220)); dot("$C$", C, dir(320)); dot("$H_A$", Ha, dir(240)); dot("$N$", N, dir(230)); dot("$D$", D, dir(315)); dot("$E$", E, dir(150)); dot("$M$", M, dir(180)); dot("$H_B$", Hb, dir(30)); dot("$K$", K, dir(330)); dot("$H$", H, dir(60)); [/asy][/asy] Let $H$ be the orthocenter, $H_A = \overline{AH} \cap (ABC)$, and $H_B = \overline{BH} \cap (ABC)$. Claim: Let $K \in (ABC)$ be such that $\overline{AK} \parallel \overline{DE}$. Then $\overline{H_BK} \perp \overline{BC}$. Proof. We have $$\measuredangle KAH_A = \measuredangle EDA = \measuredangle ABH = \measuredangle ABH_B = \measuredangle AH_AH_B$$ which means $AH_AKH_B$ is an isosceles trapezoid, implying $\overline{H_BK} \perp \overline{BC}$. $\square$ Now, from $$-1 = (DE;M\infty_{DE}) \overset{A}= (H_AB;NK) \overset{H_B}= (H_A,H;\overline{H_BN} \, \cap \, \overline{AH}, \infty_{AH})$$ we deduce $N$, $D$, $H_B$ are collinear. It follows that $$\measuredangle DNM = \measuredangle H_BNA = \measuredangle HBA = 90^\circ - \measuredangle BAC = 90^\circ - \measuredangle EDB = 90^\circ - \measuredangle MDB,$$ which implies the conclusion.

Let's hope I didn't make any mistakes We only need to show that $\measuredangle DNM = 90-\measuredangle C$, i.e. $N,D,B'$ collinear where $B'$ is the second intersection of $BH$ and $(ABC)$. Let $BM \cap (ABC)=\{B, B_1\}$. We know that $BB_1$ is symmedian in $ABC$. Then, reflection of $B_1$ wrt $AC$ is $B-humpty$ point. So, $X,Y,B',B_1$ cyclic, where $X$ is the midpoint of $AC$ and $Y$ is the foot of altitude from $B$ to $AC$, because $$\angle XYB'= 90^{\circ}=\angle HM_BX=\angle XB_1B'$$ ($M_B$ is $B-humpty$ point). Let $B'B_1 \cap AC = T$. Then $$TX\cdot TY= TB' \cdot TB_1 = TA \cdot TC$$ which means that $(A,C;Y,T)=-1$, so $DE$ passes through $T$. Finally, Pascal Thorem on $CANB'B_1B$ shows that $N,B',D$ collinear.