Let $M$ be the other midarc $AC$ and $N’$ be the reflection of $N$ over $AC$. Let $BM$ intersects $(BON)$ again at $M_2$. By spiral similarity, we have $NXM_2Y\sim NPMQ$. Let $NM_2$ intersects $XY$ at $N_2’$. We aim to prove $NXM_2N_2’Y\sim NPMN’Q$. That is, it’s enough to show $N_2’N’\parallel BM$, which is easy by angle chasing.

Let $N'$ be the image of reflecting $N$ across $AC$. By spiral similarity, we observe that $X$ is the circumcenter of $\triangle NPN'$. Analogously, $Y$ is the circumcenter of $\triangle NQN'$. Hence, $$\angle NN'P=\frac{\pi}{2}+\frac{\angle NXB}{2}$$and $$\angle NN'Q=\frac{\pi}{2}-\frac{\angle NYB}{2}$$. Since $\angle NXB=\angle NYB$, we are done.

$PQ \parallel BN$ ($\star$) (I proved it with congruent triangles leading to $PN=BQ$. It seem, though, like more straightforward way might exist. Butterfly, maybe?). Now, homothety ($B; k=0.5$) sends $PQ$ to Simson line of $O$ wrt $\triangle BXY$, that goes through midpoint of $AC$. Another homothety ($N, k=2$) sends this Simson line back to $PQ$ (due to ($\star$)) and midpoint of $AC$ to $N'$.

Note that $\angle BXN = \angle BON = 2\angle BPN$ so $NX = XP$ and in the same way $NY = YQ$. Now consider $Z$ the intersection of the circle with center $X$ and radius $XN$ and the circle with center $Y$ and radius $YN$. Note that by angle chasing $P, Z, Q$ are collinear and $NZ \perp XY$ since is the radical axis , so $N, O, Z$ are collinear. Since $NX=XZ$ and $NY=YZ$, then $Z$ is the reflection of $N$ with respect to line $AC$

Nice and easy. Here's my solution: Let $M$ be the reflection of $N$ in $O$. Note that by Reim's Theorem, $MP \parallel OX$ and $MQ \parallel OY$. Keep $A,C,O,N$ fixed and animate $X$ on line $AC$ ($B$ is also variable). Then $X \mapsto Y$ is projective, say by inversion at $O$. Also, $$X \mapsto \infty_{OX} \mapsto M\infty_{OX} \cap \odot (O,OA)=P$$is projective. Similarly, $X \mapsto Y \mapsto Q$ is also a projective map. And, if $N'$ is the reflection of $N$ in $AC$ ($N'$ is fixed), then $P \mapsto N'P \cap \odot (O,OA)$ is also projective. So it suffices to show the given result for 3 positions of $X$. For two positions, take $X$ to be the points of intersection of $\odot (ON)$ with $AC$, in which case $P$ and $Q$ become reflection of $N$ in $X$ and $Y$. For the third, take $X$ as the midpoint of $AC$, which is an obvious degenerate case. Hence, done. $\blacksquare$

Let $T$ be the point of intersection of $BB$ and $AC$ and $PQ$ intersect $AC$ at $F$. By Desargues's Involution theorem on quadrilateral $BBAC$ and line $AC$ we have $TBMF$ is cyclic. We need to prove $\angle PFA = \angle AFM$. Since $M$ is a Miquel point of quadrilateral $PXYQ$ we have $\angle PFA = \angle XMP$. And it easy to see that $\angle AFM = \angle MPB$ so we need to prove that $\triangle MPX$ is isosceles or $X$ lies on the angle bisector of $MOP$(since $\triangle MOP$ is isosceles) but this is obvious because $\angle XOM = \angle XBM = \frac {1}{2} \angle POM$ so we are done.

UlanKZ wrote: In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear. Solution. Let $NX,NY$ meet $\omega$ again at $P',Q'$ respectively and $N'$ be the reflection of $N$ in $AC$.Let $NO\cap P'Q' = N_1$. From shooting lemma, $NX\cdot NP' = NA^2 = NY\cdot NQ'$ so $X,Y,Q',P'$ are concyclic. We have \[\angle NOY = \angle NXY = \angle P'Q'N = \angle N_1Q'Y,\]so $N_1,O,Y,Q'$ are concyclic and similarly $N_1,O,X,P'$ are concyclic. This gives us \[\angle N_1XY = \angle N_1XO +\angle OXY= \angle N_1P'O + \angle ONY = \angle OQ'N_1 + \angle NQ'O = \angle NQ'N_1 = \angle YXN\]and similarly $\angle XYN = \angle N_1YX$. This gives $N_1 = N'$ or $N'$ lies on $P'Q'$. From Reim's, $PQ'||P'Q||AC$ and as $N'$ lies on perpendicular bisector of $AC$ and $N'$ lies on $P'Q'$, so $N'$ lies on $PQ$ and we are done $~\square$

Trivial! UlanKZ wrote: In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear. We begin by noticing that $\angle XNO = \angle PBO =\angle XPO \implies XN = XP$. Similarly $YN = YQ$. Let $N_1$ be the reflection of $N$ with respect to $AC \implies X$ is the circumcenter of $\triangle NPN_1$ and $Y$ is the circumcenter of $\triangle NQN_1$. Therefore $$ 2\cdot (180^{\circ} - \angle NN_1P) = \angle NXP \overset{\text{Spiral} \text{Similarity}}{=} \angle NYQ = 2 \cdot \angle NN_1Q \iff N_1 \in \overline{PQ}$$

UlanKZ wrote: In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear. Note that $\measuredangle{PXO}=\measuredangle{BNO}=\measuredangle{OBN}=\measuredangle{OXN}\implies P$ is the reflection of $N$ about $OX$.Similarly we have $Q$ is reflection of $N$ about $OY$.Now in cyclic quadrilateral $\odot{OXYN}$ the steiner line of $N$ wrt $\triangle OXY$ is precisely $PQ$ , thus reflection of $N$ wrt $XY=AC$ lies on $PQ$.$\blacksquare$ @below my proof works for this also

We don't need $N$ is the midpoint of arc $ABC$. In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is any point on arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear.

buratinogigle wrote: We don't need $N$ is the midpoint of arc $ABC$. In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is any point on arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear. Better late than never, $$\angle YQN = \angle BQN = \frac 12 \angle BON = \frac 12 \angle BYN = \frac 12 (\angle YQN + \angle YNQ) = \angle YNQ$$Thus $YQ = YN$ similarly $XP = XN$, by spiral similarity $N$ is the spiral center sending $YX \to QP$, let $YX \cap QP = D$ so $D \in (NXP) , (NYQ)$ and since $X,Y$ are the midpoints of arcs $NP , NQ$ respectively, thus reflection of $P,Q$ lie on the line $ND$.

Let $M,N'$ be the midpoint of $AC$ and the reflection of $N$ through $M$ respectively. Let $X',Y'$ be the reflection of $X,Y$ through $M$ respectively. Note that by the generalization of Butterfly's theorem , $N,Y',P$ are collinear and similarly $N,X',Q$ are collinear. But by considering an inversion centered at $N$ and radius $NA=NC$, it's enough to prove that the 4 points $N,O,NP \cap AC =X', NQ \cap AC=Y'$ lie on a same circle ; since $O$ is the inverse of $N'$ under such inversion. But this is clear since $(NX'Y')$ is in fact the reflection of $(BON)$ across $ON$ and we're done.