Serbian Mathematical Olympiad

Solution:

Let us denote by $\gamma$ the inscribed circle of $\triangle ABC$, and by $\gamma_{a}$ and $I_{a}$ the excircle opposite $A$ and its center. The point $E$ symmetric to the point $D$ with respect to $M$ is the point of tangency of $\gamma_{a}$ with the side $BC$. The perpendicular $\ell_{1}$ from $D$ to $MI$ is the radical axis of the circle $\gamma$ and the circle $\omega$ with diameter $DE$, while the perpendicular $\ell_{2}$ from $AI$ is the radical axis of the circles $\gamma$ and $\gamma_{a}$ (because $MD=ME$). Also, the radical axis of the circles $\gamma_{a}$ and $\omega$ is the perpendicular $\ell_{3}$ from $E$ to $MI_{a}$. The lines $\ell_{1}, \ell_{2}$ and $\ell_{3}$ intersect at the radical center $S$ of the circles $\gamma, \gamma_{a}, \omega$. On the other hand, it is known (from the "Great Problem") that $MI \parallel AE$ and $MI_{a} \parallel AD$ hold, so the lines $\ell_{1}$ and $\ell_{3}$ contain the altitudes from $D$ and $E$ in the triangle $ADE$. It follows that $S$ is the orthocenter of $\triangle ADE$, and it lies on the altitude from the vertex $A$.

Second solution.

Let the perpendiculars from $M$ and $D$ to $AI$ and $MI$, respectively, intersect at the point $S$, and let $J$ be the point at which the line $MI$ intersects the altitude from $A$ in $\triangle ABC$. It suffices to prove that $AJ=ID$. Indeed, then it will follow that $AJDI$ is a parallelogram, so $MS \perp DJ$, so that $D$ is the orthocenter of the triangle $MSJ$, and from this $JS \perp MD$, i.e. $AS \perp BC$.

This is computed directly. Let us denote by $H$ and $F$, respectively, the feet of the altitude and the angle bisector from the vertex $A$, and $a=BC, b=CA, c=AB$. Then $BF=\frac{ac}{b+c}$, $BD=\frac{a-b+c}{2}$ and $BH=\frac{a^{2}-b^{2}+c^{2}}{2a}$, so $FH=BF-BH=\frac{|b-c|((b+c)^{2}-a^{2})}{2a(b+c)}$, $FD=BF-BD=\frac{|b-c|(b+c-a)}{2(b+c)}$ and finally $\frac{AJ}{AH}=\frac{FD}{FH}=\frac{a}{a+b+c}=\frac{ID}{AH}$.