Belarusian Mathematical Olympiad

Let $\Gamma$ be incircle of the triangle $ABC$. Let $K$ be the point of tangency of $\Gamma$ and the side $AB$. Let the line $CL$ meet the side $AB$ at $N$. It is easy to see that $AK=BN$ (it suffices to consider the homothety with the center $C$ which transform $\Gamma$ into excircle touching $AB$ at $N$, and use the power-point theorem). Since $M$ is the midpoint of $AB$, we have $KM=AM-AK=BM-BN=MN$. So $IM\parallel LN$. If $H,I,M$ are collinear, then $HI\parallel CL$, and since $CH\perp AB$ and $LI\perp AB$, we see that $CH\parallel LI$. Therefore, $CHIL$ is a parallelogram. It follows that $CH=LI$. But $LI=r$, where $r$ is the radius of incircle of $ABC$. Thus, $CH=r$.