Show that $\angle PAB = \angle CAQ$

Let $M, N$ be two intersection points of the perpendicular bisector of $BC$ with the circumcircle $(O)$ of three points $A,B,C$

WLOG, we assume further that the point $M$ is in the interior of the angle $\angle BAC$, i.e $M$ is the midpoint of the arc $BC$ which doesn't pass through the point $A$
Also, WLOG, we assume $P$ near $A$ than $Q$.

By some argular arguments, we can see that: $$\angle MBP= \angle MBQ $$ Indeed, by symmetry, we have

• $\angle QBC= \angle QCB = \angle ACQ- \angle ACB= 180^o-\angle ABP -\angle ACB $
• $\angle MBQ= \angle QBC- \angle MBC =\angle QBC- \frac{1}{2} \angle BAC$ $= \frac{1}{2}( \angle QBC +\angle QBC-\angle BAC )=\frac{1}{2}( \angle QBC +180^o-\angle ABP -\angle ACB-\angle BAC )$ $= \frac{1}{2}(\angle QBC+\angle ABC-\angle ABP)=\frac{1}{2}(\angle QBA-\angle ABP)$ $=\frac{1}{2}\angle QBP$

Besides, $BM \perp BN$, Thus $(O)$ is the Apollonian circle for two points $P,Q$ and the ratio $\frac{MP}{MQ}$
AS $(O)$ passes through $A$, which makes $AM$ be the bisector of $\angle PAQ$
Hence, the conclusion $\square$

P/s: Oh my rusty brain cells of Olympiad toolbox.

Let $R$ be the intersection of $PQ$ with $AC$ and let $O$ be the circumcenter of $ABC$. Clearly, $O$ lies on the line $PQR$.

Clearly $\angle PBA = \angle PBO + \angle OBA$ and $\pi - \angle ACQ = \angle CQR + \angle QRC$. Since $\angle OBA = \frac \pi 2 - \angle ACB = \angle QRC$ and $\angle PBA + \angle ACQ = \pi$ (by assumption), it follows that $\angle PBO = \angle CQR = \angle OQB$. This proves that $\triangle OPB \sim \triangle OBQ$. This yields $OB^2=OP\cdot OQ$. As $O$ is the circumcenter of $ABC$, we have $OB=OA$ and therefore $OA^2=OP\cdot OQ$. This implies that $\triangle OPA \sim \triangle OAQ$, hence $\angle OAP = \angle OQA$. Finally, $$\angle PAB = \angle OAB - \angle OAP = \angle QRC - \angle OQA = \angle QAC.$$

Drawing the circumcircle about triangle $ABC$, since $P$ is on the perpendicular bisector of chord $BC$, $P$ lies on the diameter. Extend $BP$ to $F$, join $FC$ and extend it to meet the diameter's extension at $Q$. Join $QA$, crossing the circle at $D$. We want to show$$\angle PAB=\angle CAQ$$

Note that since $\angle ACF=\angle ABP$, the supplement of $\angle ACQ$, drawing and extending $FC$ from $F$ on the circumcircle determines point $Q$.

Join $DP$ extended to $K$, join $PC$, and $KQ$ crossing the circle at $E$. Since $QP$, the perpendicular bisector of $BC$, bisects $\angle BQF$, then$$\frac{BP}{BQ}=\frac{FP}{FQ}$$And by congruent triangles$$\frac{BP}{BQ}=\frac{CP}{CQ}$$Therefore points $B$, $F$, $C$ lie on an Apollonian circle with $P$, $Q$ as "foci", and we have$$\frac{DP}{DQ}=\frac{KP}{KQ}$$and hence$$\angle DQP=\angle KQP$$Therefore $AH$, $KH$ are equal arcs and, by symmetry, $AP$ extended is concurrent with $KQ$ at $E$, and makes $EG$, $DG$ likewise equal arcs. And since $G$ is thus the midpoint of arcs $DE$ and $BC$, it follows that$$\angle BAE=\angle CAD$$that is$$\angle PAB=\angle CAQ$$