Let the orthocenter be $H$. It is well known that $(AHC)$ is just the reflection of $(ABC)$ across $AC$ and so $A'$ is just $(AHC) \cap BH_b$ and similarly, $B' = (BHC) \cap AH_a$. Now, its just PoP, $A'H_a.H_aH = CH_a.H_aB = QH_a.H_a.P$ and so $A'B'PQ$ is cyclic

bigant146 wrote: In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic. What... We claim that $A'QHPB'$ is cyclic. let $AH_A$ intersects $(ABC)$ at $A_1$. By POP, we have \[ A'H_A \cdot H_A H = AH_A \cdot H_A A_1 = QH_A \cdot H_A P \]Hence, $(A' Q HP)$ is cyclic. Do this similarly, and we get $(BP'HQ)$ is cyclic, and we are done.

bigant146 wrote: In an acute triangle $ABC$ let $AH_a$ and $BH_b$ be altitudes. Let $H_aH_b$ intersect the circumcircle of $ABC$ at $P$ and $Q$. Let $A'$ be the reflection of $A$ in $BC$, and let $B'$ be the reflection of $B$ in $CA$. Prove that $A', B'$, $P$, $Q$ are concyclic. Note that the orthocenter $H$ lies on $(A'B'PQ)$ , then power of a point will finish it

Let AHa and BHb meet circumcircle at A1 and B1. H is orthocenter. We will prove B'PHQ and A'QHP are cyclic. B'Hb = Bhb , B1Hb = HHb ---> B'Hb . HbH = B1Hb . HbB = PHb . HbQ ---> B'PHQ is cyclic. We can prove A'QHP is cyclic the same way so we have proved A'QHPB' is cyclic and we're Done.

$AH_a$ and $(ABC)$ intersection $X$. $$BH_a \cdot CH_a = H_aX \cdot AH_a = HH_a \cdot H_a A'= PH_a \cdot QH_a$$$PHQA'$ - cyclic.Similarity $PQHB'$ cyclic. $PQA'B'$ - cyclic quadrateral.$\blacksquare$.