Let $D, E$ are feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ $\triangle AKH \sim \triangle ABC \sim \triangle HBL$ and $\triangle DEK \sim \triangle AHK$, $\triangle QPL \sim \triangle BHL$. So $DE \parallel PQ$ Easy to show, that $AE= AC \cos^2 \angle BAC,EC=AC \sin^2 \angle BAC$ and $PC=BC \sin^2 \angle ABC$ $\frac{EC}{PC}=\frac{AC \sin^2 \angle BAC}{BC \sin^2 \angle ABC}= \frac{BC}{AC}=\frac{BL}{HL}=\frac{LQ}{LP}=\frac{EK}{KD}$ so $P,Q,D,E$ lies on same line.

Let $M, N$ are feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ respectively. Since $HC$ is a radical axis of the circles, which are circumscribed around $ANMH$ and $HQPB$, and $C$ belongs $HC$ we get: $$ CP*CB=CN*CA $$So, the quadrilateral $ANPB$ is cyclic and $\angle BAC = \angle NPC$. We know that $\triangle AKH \sim \triangle ACB \sim \triangle HLB$ and we get $\angle BAC = \angle NMK=\angle QPL=\angle QPC$. So $\angle NPC = \angle QPC$ and we get that $P,Q,N$ lies on the same line. Similarly we can prove that $P$ belongs $MN$. And since $Q$ belongs $NP$ we are done.

feet of altitude from miquel point are collinear RMO 2006 P1

Let AE, HD perp KE, KH and F is orthocentre hence HEQF cylic so easy angle chase so done

Let the foots of perpendicular from $H$ and $A$ to $AK$ and $HK$ are $Z$ and $Y$, respectively. Then $(H Z C P)$ is cyclic, then we have $\angle HPZ=\angle HCA=90- \angle A=90-\angle BHQ=\angle HBQ=\angle HPQ$, which means that $Z$ lies on $PQ$, similarly we can prove that $Y$ lies on $PQ$ (let $BQ$ and $AY$ intersect at $X$, use that $(H Q X Y)$ cyclic).