TITLE: Show that $\angle PAB = \angle CAQ$
QUESTION [7 upvotes]: 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 $B$

REPLY [2 votes]: 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!$$\tag*{$\blacksquare$}$$