Intersection of perpendicular bisector and circle in a triangle. 
In $\triangle ABC$, $\measuredangle C=2\measuredangle B$. $P$ is a point in the interior of $\triangle ABC$, that lies on the perpendicular bisector of $BC$ and the circle centred at $A$ that passes through $C$.
Show that $\angle ABP=30^{\circ}$.


Using the definition of point $P$, $$PB=PC ~~\text{and}~~ AP=AC.$$
Let $\angle CBP=\alpha$ and $\angle ABP=\beta$. Simple angle chasing leads to $\angle CAP=180^{\circ}-2\alpha-4\beta\;$ and $\;\angle BAP=\beta-\alpha.$

I am trying to prove, $~2\angle BAP=\angle CAP~$ since this would imply $\beta=30^{\circ}$.
There is probably a smart construction that I am unable to find. Also, dropping perpendiculars from $A$ to  $CP$ and $P$ to $AB$ looks like a good attempt to reach the desired result.


Please note that $DE$ is perp bisector of $BC$ and therefore $CE$ is angle bisector of $\angle C$.
Extend $CE$ to $F$. Now $\angle PAF = 2 \angle PCF = 2 \beta$. So it follows that $\angle EAF = \alpha + \beta$ and as $\angle AEF = 180^\circ - 2 (\alpha + \beta), \angle AFP = \alpha + \beta = \angle EAF$.
As $AF \parallel BC$, $DE$ is also the perp bisector of $AF$. So we have $AP = PF$ and we conclude that $\triangle APF$ is equilateral, leading to $~2 \beta = 60^\circ$ i.e $~\beta = 30^\circ$.
