SILK ROAD MATHEMATICS COMPETITION XVII

Suppose $PQ$ intersects $AK$ and $HK$ at $R$ and $S$, respectively. From the fact that $P$ and $Q$ lie on a circle with diameter $HB$ and from $HL\parallel AC$, it follows that $\angle LPQ=\angle BHQ=\angle BAR$, i.e. the points $A,R,P$ and $B$ lie on the same circle. Therefore, $CP\cdot CB=CR\cdot CA$. Also from the right triangle $BCH$, we have $C{H}^2=CP\cdot CB$, implying $C{H}^2=CR\cdot CA$ and hence $HR\perp AC$.



It remains to prove that $AS\perp HK$. From the fact that $A,R,P$ and $B$ lie on one circle and $HK\parallel BC$ it follows that $\angle CRP=\angle CBA=\angle KHA$, i.e. the points $A,R,S$ and $H$ lie on the same circle. Since $\angle ARH=90^{\circ }$, we have $\angle ASH=90^{\circ }$ and $AS\perp HK$.