Let the circle center at $X$ pass through $P$ intersects $XY$ at $Z$ so that $Z$ lies on the same side with $Y$ w.r.t. $X$ on $XY$. Similarly, let the circle center at $Y$ pass through $Q$ intersects $XY$ at $T$ so that $T$ lies on the same side with $X$ w.r.t. $Y$ on $XY$. Let the homothety center at $C$ ratio $2$ sends $X,Y,O$ to $X',Y',C'$ respectively. Also let $CH$ intersects $(ABC)$ again at $K$. First, we have $XY\parallel X'HY'\parallel BC\parallel KC'$. Since the midpoint of $HK$ lies on $BC$, so does the midpoint of $Y'C'$. This gives $\angle{C'Y'B'}=\angle{ABY'}\implies \angle{OYB}=\angle{ABC}\implies QY=QB=YT$. Similarly, $PX=PA=XZ$. Hence, both $YTQB$ and $XZPA$ are parallelograms. So the midpoint of $BT$ lies on $YQ$ and the midpoint of $AZ$ lies on $XP$. On the other hand, from $XP+YQ=AB+XY$ we get that $TZ=XZ+YT-XY=AB$. Since $TZ\parallel AB$, $TZBA$ is a parallelogram, which means the midpoints of $BT$ and $AZ$ coincide. Hence, that said midpoint must be $O$. This implies $$\text{dist} (O,XY)=\text{dist} (O,AB)=\frac{CH}{2}=\text{dist} (H,XY)\implies OH\parallel XY\implies \angle{OHC}=90^{\circ}.$$

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.055462497926783, xmax = 23.44240141972585, ymin = -17.55529519625766, ymax = 20.243965550995654; /* image dimensions */ /* draw figures */draw(circle((-0.0027096766717737307,-0.4898061669070093), 9.604341057068408), linewidth(0.8)); draw((-6.298536593692774,6.763188426459496)--(-8.97405320845658,-3.918853027944755), linewidth(0.8)); draw((-8.97405320845658,-3.918853027944755)--(8.852168042597256,-4.209282973836018), linewidth(0.8)); draw((8.852168042597256,-4.209282973836018)--(-6.298536593692774,6.763188426459496), linewidth(0.8)); draw((-6.298536593692774,6.763188426459496)--(-6.4150024062085516,-0.385335241507258), linewidth(0.8)); draw((-7.1903754652807095,3.2025079416580793)--(7.184956111937159,-4.182120275472096), linewidth(0.8)); draw((-1.2483017149294302,3.105697959694324)--(1.2428823615858826,-4.085310293508344), linewidth(0.8)); draw((-6.4150024062085516,-0.385335241507258)--(-0.0027096766717737307,-0.4898061669070093), linewidth(0.8)); draw((-6.4150024062085516,-0.385335241507258)--(-6.47323531246644,-3.959597075490634), linewidth(0.8)); /* dots and labels */dot((-0.0027096766717737307,-0.4898061669070093),dotstyle); label("$O$", (0.15678087500440058,-0.09107978771657348), NE * labelscalefactor); dot((-8.97405320845658,-3.918853027944755),dotstyle); label("$A$", (-8.814562656780405,-3.5201266487543212), NE * labelscalefactor); dot((8.852168042597256,-4.209282973836018),dotstyle); label("$B$", (9.008506493032076,-3.7992351141876264), NE * labelscalefactor); dot((-6.4150024062085516,-0.385335241507258),linewidth(4pt) + dotstyle); label("$H$", (-6.262713829961616,-0.051207149797529904), NE * labelscalefactor); dot((-6.298536593692774,6.763188426459496),linewidth(4pt) + dotstyle); label("$C$", (-6.143095916204485,7.085995037711271), NE * labelscalefactor); dot((-7.1903754652807095,3.2025079416580793),linewidth(4pt) + dotstyle); label("$X$", (-7.857619346723359,3.6569481766735232), NE * labelscalefactor); dot((-1.2483017149294302,3.105697959694324),linewidth(4pt) + dotstyle); label("$Y$", (-1.0792709004859504,3.4177123491592614), NE * labelscalefactor); dot((7.184956111937159,-4.182120275472096),linewidth(4pt) + dotstyle); label("$P$", (7.333855700432245,-3.8789803900257134), NE * labelscalefactor); dot((1.2428823615858826,-4.085310293508344),linewidth(4pt) + dotstyle); label("$Q$", (1.3928326504947515,-3.7593624762685827), NE * labelscalefactor); dot((-6.47323531246644,-3.959597075490634),linewidth(4pt) + dotstyle); label("$E$", (-6.302586467880659,-3.639744562511452), NE * labelscalefactor); dot((-0.060942582929661704,-4.064068000890386),linewidth(4pt) + dotstyle); label("$F$", (0.11690823708535701,-3.7593624762685827), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Notice that $\angle XHC=\angle XCH=\angle OCB=\angle OBC$, hence $\triangle CXH\sim\triangle COB$. By spiral similarity, $$\triangle CXO\sim\triangle CHB$$Hence $\angle AXP=180^{\circ}-\angle CHB=\angle CAB=\angle XAP$, so $XP=PA$, similarly $YQ=QB$. Using directed length, we obtain $$AB+XY=XP+QY=AP+QB=AB+QB+BP=AB+QP$$therefore $XY=QP$. Let the projection of $H$ and $O$ on $AB$ be $E,F$, let $XY$ and $CH$ intersect at $G$, then $GE=2OF=CH$, which implies $HE=CG=OF$, so $HEFO$ is a rectangle, from which we have $\angle OHC=90^{\circ}$.

Here's another way to prove PA=PX. The rest of problem is easy from that point.

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mathaddiction wrote: let $XY$ and $CH$ intersect at $G$, then $GE=2OF=CH$ can you explain this part please.