[asy][asy]size(5cm); pair B=(-1,0),C=(1,0),A=(0,3),X,Q,P,Y; X=(-0.4,0); P=(A-B)*(X-B)/(C-B)+B; Q=A+X-P; Y=reflect(P,Q)*X; D(MP("A",A,N)--MP("P",P,E)--MP("B",B,S)--MP("X",X,S)--MP("C",C,S)--MP("Q",Q,E)--A); D(X--P--Q--X,magenta); D(X--MP("Y",Y,W)--P--Y--Q,magenta+dashed); D(circumcircle(A,B,C),green); D(B--Y--C,heavycyan); dot(A^^B^^C^^X^^P^^Q^^Y); [/asy][/asy] Assume the above configuration WLOG. We have $\frac{YP}{PB}=\frac{XP}{PB}=\frac{XQ}{QC}=\frac{YQ}{QC}.$ Also $\angle YPB=\angle YQC$ by angle chase, so $\triangle YPB\sim\triangle YQC$. So $Y$ is the center of the spiral similarity taking $PB$ to $QC$, thus the conclusion follows. $\blacksquare$ Remark. This strongly smells of APMO 2016 #1.

Dear Mathlinkers, you can see http://jl.ayme.pagesperso-orange.fr/Docs/Simplicity%201.pdf p. 27 Sincerely Jean-Louis

Solution: Let $Y $ be the refection of $X $ about $PQ $. Join $YA $, $YC $, $YQ $, $YP $ and $YB $. Simple angle chasing proves that $AYPQ $ is cyclic. So, $\angle YQC $ = $\angle YPB $... ($1$) Also, $\Delta BPX $ ~ $\Delta CQX $ (by angle chase). So, $\frac {PB}{CQ} $ = $\frac {PX}{QX} $ = $\frac {PY}{QY} $.... ($2$) ($1$) and ($2$) implies that $Y$ is the center of the spiral similarity that takes $PQ $ to $BC $. Hence, $\angle BCY $ = $\angle PQY $ = $\angle BAY $, and the result follows.

Let $Q'$ on $AB$ such that $AQ'=AQ=PX=PB$. Then $(ABC)$ and $(PQQ')$ are concentric, so the perp. bisector of $PQ$ passes through $O$ the circumcenter of $(ABC)$. Thus, reflection about the perp. bisector of $PQ$ preserves $(ABC)$. Since $Y$ is the reflection of $X$ about $PQ$, it is also the reflection of $A$ about the perp. bisector of $PQ$, so it must lie on $(ABC)$. $\Box$

Similar to the above, but slightly different. Angle chasing tells us that $\angle{PYQ}=\angle{PXQ}=\angle{PAQ}$. We also have $YQ=QX=AP$, so it follows that $APQY$ is an isosceles trapezoid, thus $Y$ is the reflection of $A$ over the perpendicular bisector of $PQ$. If we prove that the perpendicular bisector of $PQ$ goes through the origin, then we would be done. This is equivalent to proving that $P,Q$ are equidistant from the centre, or that their powers are the same. However, it is obvious that $QC \times QA=PA \times PB$, proving the result.

Let $QY \cap AB =X, AC \cap PY = T$. We have $\angle{XPQ}=\angle{XQP}$ and $\angle{YPQ}=\angle{AQP}$ so $\angle{YPA}=\angle{YQA}$ and $YAQP$ is cyclic ; further $\angle{AYQ}=\angle{APQ}=\angle{YQP}$ so $AY \parallel PQ$. Now we want to show that $\angle{YAB}=\angle{YCB}=\frac{\angle{YQX}}{2}$ so it suffices to prove that $Q$ is the center of $\odot{YXC}$. Indeed, $QY=QX$ trivially and since $\angle{XQC}=A$. Further, $\angle{ACX}=90-\frac{A}{2}$ so we get $QX=QC$ as desired.

Note that $PY=PX=PB$ and $QY=QX=QC$ and by angle chase, $\measuredangle YPB=2\measuredangle QPX+\measuredangle XPB=2\measuredangle PQX+\measuredangle XQC=\measuredangle YQC$. Hence, $Y$ is the center of spiral similarity taking $\overline{PB}\to\overline{QC}$, hence $Y$ is also the center of spiral similarity taking $\overline{BC}\to\overline{PQ}$. Therefore, $Y$ lies on $(ABC)$ and $(APQ)$. $\blacksquare$