Intersection of perpendicular bisector and circle in a triangle.

Question

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 $\;\angle BAP=\beta-\alpha.$

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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.

Answer 1

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 $~\beta = 30^\circ$.

Answer 2

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 $\beta = 30^{\circ}.$