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.

Connect $CE$, and we may notice that $\angle ACE=\angle ECB=\angle CBA.$
Since $\angle CAE=\angle BAC,\angle ACE=\angle ABC$, we get $\triangle EAC \sim \triangle CAB.$
Thus  $AC^2=AE\cdot AB.$
Since $AC=AP$, we have $AP^2=AE\cdot AB.$
And because $\angle PAE=\angle BAP$, we got $\triangle PAE \sim \triangle BAP.$
Thus $\angle ABP=\angle APE=\beta.$
Since $\angle CPD+\angle CPA+\angle APE=180^{\circ}$,  $(90^{\circ}-\alpha)+(\alpha+2\beta)+\beta=180^{\circ}.$
So $\beta = 30^{\circ}.$
