In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.

Let $X, Y$ be the feet of the altitudes of $\triangle AKH$ emanating from the vertices $A$ and $H$ respectively. Because $HC$ is the radical axis of $(AYXH),(HQPB)$, and $C$ belongs $HC$, we obtain $CB \cdot CP=AC\cdot YC$. Hence, the quadrilateral $AYPB$ is cyclic, implying $\angle BAC = \angle YPC$. Next, because $\triangle AKH \sim \triangle ACB \sim \triangle HLB$, we have $\angle BAC = \angle YMK=\angle QPL=\angle QPC$. Therefore, $\angle YPC = \angle QPC$ which implies that $P,Q,Y$ lies on the same line. In the same way, we show that $P \in MY$. And because $Q \in YP$, we have proven the desired. $\square$