Here is another solution: Claim: The points $P$, $Q$, $E$, and $F$ are concyclic Since $BFEC$ is concyclic the result follows by Reims Theorem. Let $X$ be the center of $(PQEF)$. Claim: $X$ lies on $(APF)$ $$\angle FXP=2\angle FQP=\angle PAH+\angle FCB=\angle PAF$$ Claim: $X$ lies on $(AQF)$ $$\angle QXE=2\angle QPH=\angle QAH+\angle EBC=\angle QAE$$ Claim: $AX||BC$ $$\angle XAF=\angle XPF=90^{\circ}-\angle PQF=\angle ABC$$This implies the result.

By the conditions we can easly get ${AB, AP}, {AC, AQ}$ symetric over $AH$, by Reim's theorem $QEPF$ is cyclic, now let $X'$ the center of $(QEPF)$. Claim: $X=X'$ Proof: Just note that $\angle FX'P=2\angle FEP=2\angle BAH=\angle FAP$ so $X'$ lies in $(APF)$, in the same way one can get $X'$ lies in $(AQE)$ and done. Finish: Since $XF=XP$ we have that $X$ is midpoint of arc $FAP$ in $(FAP)$, now $AH$ is internal angle bisector of $\angle FAP$ therefore $\angle XAH=90$ so $XA \parallel BC$ as desired.

$\textbf{Solution 1.}$ Let $E$ and $F$ be the feet of altitudes from $B$ to $AC$ and $C$ to $AB$ respectively. Since $BCEF$ is cyclic, then by Reim's Theorem $PQEF$ is cyclic too. Let $O$ be the circumcenter of $PQEF$. We claim that $X=O$. It suffice to prove that $AOPF$ is cyclic. Using directed angles we can angle chase that $$\angle(HA,AP)=\angle(HQ,QP)=\angle(HC,CB)=\angle(FA,AH)$$So $AP$ is the reflection of $AF$ about $AH$, thus we have $$\angle(FO,OP)=2\angle(HQ,QP)=\angle(HA,AP)+\angle(FA,AH)=\angle(FA,AP)$$Hence $AOPF$ is cyclic. Similarly, $AOQE$ is cyclic too, so $X=O$. Finally we angle chase that $AO\parallel BC$: $$\angle(OA,AF)=\angle(OP,PF)=90^{\circ}-\frac{\angle(FO,OP)}{2}=90^{\circ}-\angle(HQ,QP)$$$$=90^{\circ}-\angle(FA,AH)=\angle(BC,AF)$$So $AO\parallel BC$ as desired. $\blacksquare$ $\textbf{Solution 2.}$ (Loke Zhi Kin) Just like in the previous solution we can angle chase that $AP, AQ$ is the reflection of $AF, AE$ about $AH$. To show that $AX$ is parallel to $BC$, it suffices to show that the line connecting the centers of circles $(APF)$ and $(AQE)$ is perpendicular to $BC$. Let $AX$ and $AY$ be the diameters in $(APF)$ and $(AQE)$ respectively, then it suffices to prove that $XY$ perpendicular to $BC$. Note that the lines $FX, PX, PQ$ are perpendicular to the lines $AF, AP, AH$ respectively. Then by a $90$ degree rotation of the lines, since $AF$ and $AP$ are reflections of each about across $AH$, then $FX$ and $PX$ must subtend the same angles with $PQ$. Thus $\angle XQP=\angle XPQ$, that is $X$ lies on the perpendicular bisector of $PQ$. Similar conclusion holds for $Y$ too, so $XY$ is the perpendicular bisector of $PQ$, hence perpendicular to $BC$. $\blacksquare$ $\textbf{Solution 3.}$ (Anzo Teh) As before we prove that $PQEF$ is cyclic. By Radical Axis Theorem on $(APF), (AQE), (PQEF)$, then the lines $AX, FP, EQ$ are concurrent. Let $\ell$ be the line parallel to $BC$ through $A$. We want to prove that $X\in \ell$, which is enough to prove that $\ell, FP, EQ$ are concurrent. $\textit{Claim:}$ The center of the circle $(PQEF)$ lies on $\ell$. $\textit{Proof:}$ Let $M$ be a point such that $AMPQ$ is an isosceles trapezoid with $AM$ parallel to $PQ$. Then since $HA$ is the altitude in triangle $HPQ$, $HM$ must be the diameter of the circle $(AHPQ)$, hence $MQ\perp QH$ and $MP\perp PH$. Now let $N$ be the midpoint of $AM$, then $NP=NQ$ since $AMPQ$ is an isosceles trapezoid. Moreover, from the trapezoid $AMQF$ and $AMPE$, we have $NQ=NF$ and $NP=NE$ too. Hence $N$ is the circumcenter of $(PQEF)$. Finally, $AN\parallel PQ$, so $N\in \ell$. $\square$ Now, let $Y$ be the intersection of $FP$ and $QE$, and let $Z$ be the intersection of $QP$ and $FE$. By Radical Axis Theorem on $(AHPQ), (AHEF), (PQEF)$, we have $Z$ lies on $AH$. Then by Brokard's Theorem, $YN\perp HZ$, which is $YN\perp HA$. Since $N\in \ell$ and $\ell \perp HA$, we have $Y\in \ell$ too. So the lines $\ell, FP, EQ$ are concurrent. $\blacksquare$

navi_09220114 wrote: Let $ABC$ be an acute triangle with orthocenter $H$, and let $BE$ and $CF$ be the altitudes of the triangle. Choose two points $P$ and $Q$ on rays $BH$ and $CH$ respectively, such that: $\bullet$ $PQ$ is parallel to $BC$; $\bullet$ The quadrilateral $APHQ$ is cyclic. Suppose the circumcircles of triangles $APF$ and $AQE$ meet again at $X\neq A$. Prove that $AX$ is parallel to $BC$. Proposed by Ivan Chan Kai Chin 1)Prove that line $AQ$ is the symmetrical of $AC$ with respect to $AH$ similar for $AP$. 2)Prove that $EPFQ$ lie on the same circle. 3)Prove that $X$ is the center of $EPFQ$ 4)Prove that $AX$ is // to $BC$

By applying radax theorem on $(AH), (APHQ), (PQEF)$ and $(AQE),(APF),(PQEF)$ we get $PQ, EF, AH$ and $PF, EQ, AX$ concur at say points $Y,Z$ respectively. By reim's theorem we get $\odot(PQEF)$ is cyclic, now we take $X'$ to be the center of $\odot(PQEF)$ and we prove $X'\in (APF), (AQF)$, this is true because we have \begin{align*} &\angle PX'F=2\angle PEF=\angle PAF\\ &\angle QX'F=2\angle QEF=\angle QAF \end{align*}So $X\equiv X'$. Now we use the fact that $AH$ bisects $\angle FAP$ and finish by angle chasing, $$\angle XAF=\angle XPF=90^{\circ}-\frac{\angle FXP}{2} =90-\angle FAH=\angle ABC$$So $AX\parallel BC.$