This problem is indeed Silk Road Math Competition 2018 P1

Let $T$ $\equiv$ $QS$ $\cap$ $AB$ We have: $CH^2$ = $\overline{CQ}$ . $\overline{CB}$ = $\overline{CS}$ . $\overline{CA}$ so: $A$, $B$, $Q$, $S$ lie on a circle Then: $\overline{TB}$ . $\overline{TA}$ = $\overline{TQ}$ . $\overline{TS}$ = $TH^2$ or $\dfrac{\overline{TB}}{\overline{TH}}$ = $\dfrac{\overline{TH}}{\overline{TA}}$ = $\dfrac{\overline{BH}}{\overline{HA}}$ = $\dfrac{\overline{BQ}}{\overline{HR}}$ Hence: $T$, $Q$, $R$ are collinear Similarly: $T$, $P$, $S$ are collinear Then: $T$, $Q$, $P$, $R$, $S$ are collinear

CSHP is cyclic so SPH=HCB=90-B=QAH=QPH so P,Q,R collinear and PSH=HCA=90-A=RBH=RSH so Q,R,S collinear so P,Q,R,S collinear.

$HL||AC$ and $HK||BC$ $\implies$ $BP,AR,CH$ concur at the orthocenter of $\Delta ABC$, Let it be $D$ and let $H_A,H_B$ be the foot of altitude from $A,B$ in $\Delta ABC$, then either use: Approach 1[hide]$H$ is the miquel point of the complete quadrilateral $AH_BCH_ABD$, hence done![/hide] or Approach 2

One can also prove $PQ || RS$ using angle chasing and then show one set of collinearity.

Let foot of $A,B$ in $ABC$ be $X,Y$ and so, simson line in $BDX, ADY$ completes the proof. P.S: Too easy for a TST

Let $O=KL\cap AB$ be the homothetic center of $\triangle HLB$ and $\triangle AKH$. Let $X$ be the foot of $L$ onto $AB$. Now let's use projective geometry. Clearly $LHKC$ is a paralelogram, so $KL$ bisects $HC$. Hence, by projecting thorough $L$, $(H, B; X, O)=-1$. But that means that, by the "cevians induce harmonic bundles lemma", if $LO=PQ\cap AB$ then $(H, B; X, O')=-1\implies O'=O$. Thus $O-P-Q$ are colinear points. Since clearly $O-P-S$ and $O-Q-R$ are colinear through the homothety, we are done.