can you post p4 and p5?

Well, this problem is wonderful! Here is my solution:

Let $D=AC\cap PQ$ and $H$ the orthocenter. Note that by Reim's since $HKCL$ is cyclic, $DPLH$ is cyclic as well. Again by Reim's since $DPLH$ is cyclic $QDHB$ is cyclic as well. Now angle chase to get $\angle LBA=\angle LKA=\angle LPH=\angle LDH$, so $DAHB$ Is cyclic, and since $QDHB$ is cyclic too we have that $Q, D, A, H, B$ are all concyclic, in particular $QAHB$ is cyclic, so $\angle AQB=\angle BHK=\angle BCA$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%2010.pdf p. 43... Sincerely Jean-Louis

Let $H$ be the orthocenter of $ABC$ , Its easy to see that it suffices to prove that $QAHB$ is cyclic. Let $AC$ Intersect $PQ$ at $X$ we know that $$ HXL=HPL=LKP=LBA $$therefore $XAHB$ and $PXHL$ are cyclic so $PXH+PLH=180$ and because $PL$ is parallel to $BQ$ , $PLH+LBQ=180$ therefore $PXH=HBQ$ therefore $XHBQ$ is cyclic , and because both $XAHB$ and $XHBQ$ are both cyclic so the pentagon $QXAHB$ is cyclic , and we are done!!

Let H be orthocenter and PQ meet AC at S. we will prove AQBHS is cyclic. step1 : PSLH and ASHB are cyclic. (1) ∠HPS = 90 = ∠SLH ---> PSLH is cyclic. ∠ABH = ∠AKL = ∠HPL = ∠HSL ---> ASHB is cyclic. step2 : QSHB is cyclic. (2) ∠QSH = ∠PLH = 180 - ∠LBQ ---> QSHB is cyclic. now from (1) and (2) we have AQBHS is cyclic so: ∠ACB = ∠KHB ∠AQB we're Done.