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, $CH = r$.