Since XIV GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN The correspondence round, 2018 ended two days ago (1st of April), I would like to ask for help about a problem. Even though I didn't participate, I tried to solve them, but I couldn't do this one. I tried angle hunting, vectors, etc. (I didn't try trigonometry) but I couldn't solve it.
Here is the problem:
The altitudes $AH_1, BH_2, CH_3$ of an acute-angled triangle $ABC$ meet at point $H$. Points $P$ and $Q$ are the reflections of $H_2$ and $H_3$ with respect to $H$. The circumcircle of triangle $PH_1Q$ meets for the second time $BH_2$ and $CH_3$ at points $R$ and $S$. Prove that $RS$ is a medial line of triangle $ABC$.
This was my proposed solution to it (It's long because I didn't want to leave out any details, also, I probably missed a quick solution):
Let $M$ be the midpoint of $AC$, $H_2H_3 \cup AH_1 = X$, $PQ \cup AH_1 = Y$, $MH_1 \cup CH_3 = Z$.
Firstly, we will show $RS \parallel BC$.
By Power of a Point, $HR \times HP = HS \times HQ$. Since $HP=HH_2$ and $HQ=HH_3$, we have $HR \times HH_2 = HS \times HH_3$, so $H_2H_3SR$ is cyclic. Since $\angle BH_2C = \angle BH_3C = 90^{\circ}$, we have $H_2H_3BC$ cyclic. Thus, since $BC$ and $H_2H_3$ are anti-parallel and $RS$ and $H_2H_3$ are anti-parallel, we have $BC \parallel RS$
Now, we will show $M$ lies on $RS$.
We have $\angle PH_1S = \angle PRS$ by angles subtended by equal chords in $(PH_1QRS)$. Furthermore, $\angle PRS = \angle H_2BC$ since $BC \parallel RS$. Finally, since $ABH_1H_2$ cyclic (from $\angle BH_1A = \angle BH_2A$), we have $\angle H_2BC = \angle H_2 A H_1$
Since $M$ is the midpoint of $AC$ and $\angle CH_1A=90^{\circ}$, we have $M$ is the centre of the circle through $CH_1A$ by semicircle theorem, so $MC=MA=MH_1$. Therefore, $\angle H_2 A H_1 = \angle MH_1H$. Chaining these equalities yields $\angle PH_1S = \angle MH_1H$, so $\angle PH_1Y = \angle SH_1Z$.
Furthermore, by angles subtended by equal chords, $\angle H_1PY = \angle H_1SZ$ in $(PH_1QRS)$, so by the equiangular criterion, $\Delta PH_1Y \sim SH_1Z$.
By angles subtended by equal chords in $H_3H_1CA$ (from $\angle CH_1A = \angle CH_3A$), we have $\angle H_3H_1X = \angle ZCM$. Also, by the existence of the nine-point circle through $H_1, M, H_2, H_3$, we have $\angle H_1H_3D = \angle ZMC$. Thus, by the equiangular criterion, $\Delta H_1H_3X \sim CMZ$. Thus, $\frac{H_1H_3}{H_3X} = \frac{CM}{MZ}$.
From angles subtended by equal chords in $BH_1HH_3$ (from $\angle HH_3B + \angle HH_1B = 180^{\circ}$, $BH_1H_2A$ (from before) and $AH_3HH_2$ (from $\angle AH_2H + \angle AH_3H = 180^{\circ}$) respective, we have $\angle HH_3H_1 = \angle HBH_1 = \angle HAH_2 = \angle HH_3H_2$, so $H_3H$ bisects $\angle H_1H_3H_2$. Thus, by angle bisector theorem, $\frac{H_3H_1}{H_3X} = \frac{H_1H}{HX}$.
Furthermore, since $HH_2 = HP$ and $HH_3 = HQ$, by symmetry $HX = HY$. Thus, $\frac{CM}{MZ} = \frac{H_1M}{MZ} = \frac{H_1H}{HX} = \frac{H_1H}{HY}$, so we have $\frac{H_1M}{MZ} = \frac{H_1H}{HY}$. From the similarity, $\Delta PH_1Y \sim SH_1Z$, we have $\frac{H_1P}{H_1Y} = \frac{H_1S}{H_1Z}$, so we have $\Delta PH_1H \sim SH_1M$ by SAS.
Finally, we have $\angle RSH_1 = \angle HPH_1$ by angles subtended by equal chords in $(PH_1QRS)$ and $\angle HPH_1 = \angle MSH_1$ from the similarity. Thus, $\angle RSH_1 = \angle MSH_1$, so $R,S,M$ are collinear. Thus, $RS$ is the medial line of $ABC$, parallel to $BC$.