Show that $\angle PAB = \angle CAQ$ Given a triangle $\triangle ABC$ with a point $P$ in the perpendicular bisector of $BC$. Take point $Q$ also in the bisector of $BC$ s.t. $\angle ABP + \angle ACQ =180^o$ and such that $P$ and $Q$ are in the interior of $\angle BAC$.
Prove that $\angle PAB = \angle CAQ $

I tried to find some circles in this configuration. Tried to find some interesting locus and I failed so I went for trigonometry. This problem is equivalent to show that $\frac{BP}{AP} = \frac{BQ}{AQ}$... wait this looks like an apolonian circle.
Ok it looks like $CQ$ is tangent to the apollonian circle that goes throught $P$ and $Q$ EDIT: it is not tangent.
Can we work it with coaxial systems? Tha apollonian circle through fixed points $AB$ form a coaxial system orthogonal to the system of circles that pass throught $A$ and $B$
 A: 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,
$$\angle PAB = \angle OAB - \angle OAP = \angle QRC - \angle OQA = \angle QAC.$$

A: 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$$\angle PAB=\angle CAQ$$
A: 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.
A: We use @Edward Porcella's diagram!
Let $PQ\cap\odot(ABC):=\{G,H\}$ and hence,
$$ \angle ABP+\angle ACQ = 180=\underbrace{\angle ACG}_{\angle ACQ-\angle GCQ} +\underbrace{\angle ABG}_{\angle ABP+\angle PBG}$$$$\implies \angle PBG=\angle GCQ=\angle GBQ $$
Hence, $BG$ and $BH$ are internal and external angle bisectors of $\angle QBP\implies \odot(ABC)$ is $B$-Apollonian Circle of $\triangle PBQ$ and as $A$ lies on this circle, we get $AG$ bisects $\angle QAP$ and we are done!$$\tag*{$\blacksquare$}$$
