Suppose that $P$ is $A$ - Humpty point of $\triangle ABC$ then $P$ is second intersection of $(D, DA)$ and $(E, EA)$. So $DE$ must passes through midpoint of $AH$ with $H$ is orthocenter of $\triangle ABC$. From this, we have $MN$ passes through $H$

Let $H$ be the orthocentre of $ABC$. Use complex numbers with $(ABC)=\mathbb{S}^1$. Then $d=t(a+b)$ for some $t\in \mathbb{R}$, so$$\overline d=t(\overline a+\overline b)=\frac{t(a+b)}{ab}=\frac{d}{ab}.$$Hence\begin{align*}\frac{d-b}{c-b} & =-\frac{d-b}{b-c} \\ & =\overline{\left (\frac{d-b}{b-c}\right )} \\ & =\frac{\overline d-\overline b}{\overline b-\overline c} \\ & =\frac{\frac{d}{ab}-\frac{1}{b}}{\frac{1}{b}-\frac{1}{c}} \\ & =\frac{\frac{d-a}{ab}}{\frac{c-b}{bc}} \\ & =\frac{c}{a}\frac{d-a}{c-b}, \end{align*}so $d=a\frac{c-b}{c-a}$, and therefore $m=2d-a=a\frac{a-2b+c}{c-a}$. We then have\begin{align*}m-h & =a\frac{a-2b+c}{c-a}-(a+b+c) \\ & =\frac{a(a-2b+c)-(c-a)(a+b+c)}{c-a} \\ & =\frac{2a^2-c^2-bc+ca-ab}{c-a} \\ & =\frac{(c+a)(2a-b-c)}{c-a}. \end{align*}Analogously, we have$$h-n=\frac{(a+b)(2a-b-c)}{a-b}.$$Hence$$\frac{m-h}{h-n}=\frac{\frac{(c+a)(2a-b-c)}{c-a}}{\frac{(a+b)(2a-b-c)}{a-b}}=\frac{c+a}{c-a}\frac{a-b}{a+b},$$but$$\overline{\left (\frac{c+a}{c-a}\frac{a-b}{a+b}\right )}=\frac{\frac{1}{c}+\frac{1}{a}}{\frac{1}{c}-\frac{1}{a}}\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{c+a}{a-c}\frac{b-a}{a+b}=\frac{c+a}{c-a}\frac{a-b}{a+b}.$$Hence $\frac{m-h}{h-n}\in \mathbb{R}$, and therefore $M,H,N$ are collinear.

Now for the actual solution. Let $H$ be the orthocentre of $ABC$. Let the line through $A$ parallel to $CH$ intersect $BD$ at $F$ and let the line through $A$ parallel to $BH$ intersect $CE$ at $G$. Let $I=BH\cap AF$ and $J=CH\cap AG$. Then $D$ is the midpoint of $BF$ and $E$ is the midpoint of $CG$, so $ID$ and $JE$ pass through the midpoint of $AH$. But $AIHJ$ is a parallelogram, so $IJ$ passes through the midpoint of $AH$. Hence $D,E,I,J$ and the midpoint of $AH$ are collinear, and therefore $MN$ passes through $H$.

Quick trig sol that I gave in contest. Let $I$ be the intersection of $DE$ and $AH$, $P$ and $Q$ be the midpoints of $AB$ and $AC$ respectively and $O$ be the circumcenter. We need to prove that $I$ is the midpoint of $\overline{AH}$, which is equivalent to proving that $PI$ is parallel to $OE$. We have $\angle{ABP} = 90^{\circ} - \angle{B}$ so $BD = AD = \frac{c}{2\sin(B)}$ and $DP = \frac{c\cot(B)}{2}$. We also have that $OP = R\cos(C)$ so $$\frac{DP}{PO} = \frac{c\cot(B)}{2R\cos(C)} = \frac{c\cos(B)}{2R\sin(B)\cos(C)}$$Now, $\angle DAI = 180^{\circ} - 2\angle B$ and $\angle IAE = 180^{\circ} - 2\angle C$. From ratio lemma in triangle $ADE$ we have $$\frac{DI}{IE} = \frac{c\sin(2B)\sin(C)}{b\sin(2C)\sin(B)} = \frac{c\cos(B)}{b\cos(C)}$$And hence, by LOS we have $\frac{DP}{PO} = \frac{DI}{IE}$ so we are done.