Let $T=\overline{XP} \cap \overline{YQ}$, then $\overline{PQ}$ and $\overline{XY}$ are antiparallel in angle $XTY$. Also, $\overline{XY}$ and $\overline{BC}$ are antiparallel in angle $BAC$. However, bisectors of angle $BAC$ and angle $XTY$ are parallel, hence $\overline{PQ} \parallel \overline{BC}$. $\blacksquare$

Well that's probably a dumb question but do you exactly got that $APH_aQ$ is a kite?

Stats- 18 students solved it in the contest.

Suppose $X$ is closer to $H_c $. Let $H $ be the orthocenter. If we consider the inversion $\Psi $ centered at $A $ with radius $\sqrt {AH_c\cdot AB} $, then we shall see that $X $ and $Y $ remain invariant under $\Psi $. So, $AX = AY $. As $P $ and $Q $ are reflections of $X $ and $Y $ in $AB $ and $AC $ respectively, so, $AX = AP = AQ = AY $. Thus, $PXYQ $ is cyclic. But, $\angle PXY = \angle CAH$. This gives $\measuredangle PQY = \measuredangle HAC\implies \measuredangle (AC, QY) = \measuredangle (AH, PQ) $. Hence, $AH\perp PQ\implies PQ || BC $.

Lemma: With the notations given in this question we have $AX = AY$. Main Question: Suppose $X$ is closer to $H_c $. Let $H_a $ be the feet of altitude from vertex $A$. Clearly, $\angle PH_cB = \angle XH_cB = \angle C$ and $\angle QH_bC = \angle YH_bC = \angle B$. Therefore $ H_cP \cap H_bQ \equiv H_a$. Now $AX = AY \Rightarrow \angle AXH_c = \angle AYH_b \Rightarrow \angle APH_c = \angle AQH_b \Rightarrow \angle APH_a = \angle AQH_a.$ Therefore we have $\Delta APH_a \cong \Delta AQH_a \Rightarrow PH_a = QH_a \Rightarrow \angle H_aPQ = \angle H_aQP = \angle A.$ Therefore $PQ || BC$.

Pretty straightforward. Let $O$ be the circumcenter of $\triangle ABC$. $AO \perp H_bH_c$, thus, $\overline{AO}$ is the perpencular bisector of $\overline{XY}$. Therefore, $AX =AY = AP =AQ$. Note that $BC$ is antiparallel to $H_bH_c \equiv XY$ which in turn is antiparallel to $PQ$. Thus, $BC\parallel PQ$.

Here's a nice solution: Lemma 1: $AX=AY$. Proof: Note that $H_bH_c$ and $BC$ are antiparallel w.r.t $\angle B$ and $AH, AO$ are isogonal. Hence, $AO$ is the perpendicular bisector of $XY$ and so $AX=AY \implies AP=AQ$. $\square$ Lemma 2: $AH$ bisects $\angle PAQ$. Proof: Reflect $H$ over $AB, AC$. These lie on the circumcircle. Now some basic angle chasing and lemma $1$ does the job. $\square$ Hence, $AH \perp PQ$ and $AH \perp BC \implies PQ \parallel BC$. $\blacksquare$

Whats does antiparallel about any angle mean??

https://www.cut-the-knot.org/Curriculum/Geometry/Antiparallel.shtml

Here goes a long solution to this problem. Sharygin Finals 2017, Problem 9.5 wrote: Let $BH_b, CH_c$ be altitudes of an acute-angled triangle $ABC$. The line $H_bH_c$ meets the circumcircle of $ABC$ at points $X$ and $Y$. Points $P,Q$ are the reflections of $X,Y$ about $AB,AC$ respectively. Prove that $PQ \parallel BC$. Proposed by Pavel Kozhevnikov Solution:- Claim 1:- $XPQY$ is a cyclic quadrilateral. First we will prove $AX=AY$. In order to prove this if we can show $AO\perp XY$ we are done. Let $H$ be the orthocenter of $\triangle ABC$ and $H_A$ be the foot of perpendicular from $A$ to $BC$. It's well known that $AH$ and $AO$ are isogonal also $\angle AHH_C=\angle AH_BX$. Hence, by Similarity we get $AO\perp XY\implies AX=AY$. So now just notice by Congruence due to Reflections and Perpendiculars that $AX=AP=AQ=AY\implies XPQY$ is a cyclic quadrilateral. Claim 2:-$PH_C\cap QH_B=H_A$ Let $XP\cap AB=M$ and $YQ\cap AC=N$ and $H_BH_C\cap BC=X_A$ which is actually the A-Expoint of $\triangle ABC$. Let $HP\cap BC=H_A'$ So, $$-1=(X,P;M,\infty_{XP})\overset{H_C}{=}(X_A,H_A';B,C)\implies\boxed{H_A'\equiv H_A}$$Similarly if $QH_B\cap BC=H_A'''$ then $$-1=(Y,Q;N,\infty_{YQ})\overset{H_B}{=}(X_A,H_A''';C,B)\implies\boxed{H_A'''\equiv H_A}$$So $H_CP,H_BQ$ meet at $H_A$. Returning to the main proof:- Now notice that by Blanchet's Theorem or using the fact that $H$ is the incenter of the $\triangle H_AH_BH_C$ we get that $\angle CH_CH_B=\angle CH_CH_A$ and let $R$ be any point on the ray $PQ$. So, $$\angle H_BBC=\angle CH_CH_B=\angle H_AH_CC=\angle YXP=\angle YQR=\angle H_BPQ\implies\boxed{PQ\|BC.} \blacksquare$$

Sharygin Finals 2017, Problem 9.5 wrote: Let $BH_b, CH_c$ be altitudes of an acute-angled triangle $ABC$. The line $H_bH_c$ meets the circumcircle of $ABC$ at points $X$ and $Y$. Points $P,Q$ are the reflections of $X,Y$ about $AB,AC$ respectively. Prove that $PQ \parallel BC$. Proposed by Pavel Kozhevnikov Solution: Claim: $XPQY$ is cyclic with center $A$. Proof: Let $O$ be the circumcenter of $\triangle ABC$. As $AO\perp H_bH_c\equiv XY$ we have $AO$ as perpendicular bisector of $XY$. Hence, $AX=AY$. So, $$AP=AX=AY=AQ\implies XPQY \text{ is cyclic}$$as desired. $\square$ Let $XP$ and $YQ$ intersect $BC$ at $P_1$ and $Q_1$ respectively. It can be seen that $$XP_1\parallel H_cC\implies \angle XP_1B=\angle H_cCB=\angle H_cH_bB=\angle XH_bB.$$Also, as $YQ_1\parallel H_bB$ we have $\angle XYQ=\angle XH_bB$. So, $$\angle XP_1B=\angle XH_bB=\angle XYQ=\angle QPP_1\implies PQ\parallel BC.$$$\blacksquare$ [asy][asy] import olympiad; import geometry; size(300); pair A,B,C,HB,HC; A=(5,7); B=(0,0); C=(8,0); HB=foot(B,A,C); HC=foot(C,A,B); draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(B--HB); draw(C--HC); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$H_b$",HB,NE); dot("$H_c$",HC,NW); circle d=circle(A,B,C); pair X=intersectionpoints(d,line(HB,HC))[1]; dot("$X$",X,NW); pair Y=intersectionpoints(d,line(HB,HC))[0]; dot("$Y$",Y,SE); draw(X--Y); pair X1=foot(X,A,B); pair P=2X1-X; dot("$P$",P,SW); pair Y1=foot(Y,A,C); pair Q=2Y1-Y; dot("$Q$",Q,SE); pair P1=intersectionpoint(line(X,P),line(B,C)); pair Q1=intersectionpoint(line(Y,Q),line(B,C)); dot("$P_1$",P1,S); dot("$Q_1$",Q1,S); draw(X--P1); draw(Y--Q1); draw(circumcircle(X,P,Q),red+dashed); draw(P--Q,blue); draw(circumcircle(B,C,HB),dotted); pair O=circumcenter(A,B,C); draw(A--O); dot("$O$",O,S); [/asy][/asy]

A neat Complex bash: Let $(ABC)$ be the unit circle. One can easily get $A$ is mid point of arc $XY$. We solve the following stronger problem: If we take points X and Y on the unit circle such that A is mid point of arc XY then PQ//BC. Let $z$ be the complex number associated with point $Z$.By formula of reflection- $$p=a+b-\frac{ab}{x}$$$$q=a+c-\frac{ac}{y}$$Hence $$\frac{p-q}{b-c}=\frac{(b-c)-a(\frac{by-cx}{xy})}{b-c}$$So, $$\overline{(\frac{p-q}{b-c})}=\frac{\frac{(c-b)}{bc}-\frac{1}{a}(\frac{-by+cx}{bc})}{\frac{b-c}{bc}}=\frac{b-c-\frac{by-cx}{a}}{b-c}$$We use the relation of $A,X,Y$. Consider the A-antipode in the unit circle as $T$. Then $AT \perp XY$. By the lemma on perpendicular chords in complex numbers- $$xy+at=0 \implies xy+a(-a)=0 \implies xy=a^2$$Using this relation it's easy to observe $$\frac{p-q}{b-c}=\overline{(\frac{p-q}{b-c})}$$This means $PQ//BC$