Let $L$ be the common point of the symmedians of triangle $ABC$, and $BH$ be its altitude. It is known that $\angle ALH = 180^o -2\angle A$. Prove that $\angle CLH = 180^o - 2\angle C$.
Let $AX \perp BC $ and $CY \perp AB$ .
Then $AX \cap YH = \{M\}$ and $CY \cap XH =\{N\}$ will be the midpoints of $YH$ and $XH$ since $HCBY$ and $HABX$ are cyclic
combined with $K$ - symmedian point.
Using that $\angle ALH = 180 - 2\angle A$ notice that $HMLN$ is cyclic and the conclusion follows