Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle ABC so that $CL$ is a bisector and $CH$ is an altitude. Let $P,Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively. Prove that $AP \cdot BH = BQ \cdot AH$.
I. Gorodnin
We have $\angle LHC=\angle LPC=\angle LQC=90^{\circ}$ so $CPLHQ$ is cyclic.
Now by PoP we have $AP\times AC=AL\times AH$ and $BQ\times BC=BL\times BH$
so we have $AP\times BH\times AC\times BL=BQ\times AH\times BC\times AL$.
However, as $\frac{BL}{AL}=\frac{BC}{AC}$ we have $AC\times BL=BC\times AL$.
So $AP\times BH=BQ\times AH$.