Mongolian Mathematical Olympiad

Let $B{B}^{\prime }$, $C{C}^{\prime }$ be altitudes. Then points $D$, $B^{\prime }$, $C^{\prime }$ lie on the circle with diameter $AH$. Since $\angle C^{\prime }DM=\angle C^{\prime }DH=\angle C^{\prime }AH=\angle C^{\prime }CM$, points $D$, $C^{\prime }$, $M$, $C$ are cyclic. Thus we have $△C^{\prime }HM\sim △DHC\Rightarrow \frac{C^{\prime }M}{CD}=\frac{HM}{HC}\Rightarrow \frac{CD}{CH}=\frac{C^{\prime }M}{HM}$.

Similarly, we conclude that $\frac{BD}{BH}=\frac{B^{\prime }M}{HM}$. Since $C^{\prime }M=B^{\prime }M$, we get $\frac{CD}{CH}=\frac{BD}{BH}$. Bisector of the angle $\angle BDH$ is $BL$ from where follows $\frac{BD}{DH}=\frac{BL}{LH}$ and the fact that $B{L}^{\prime }$ is bisector of the angle $\angle CDH$ implies $\frac{CD}{CH}=\frac{C{L}^{\prime }}{L^{\prime }H}\Rightarrow L=L^{\prime }$.