Solution: Let angle $\angle E A B=\gamma . \mathrm{Since} \angle E A B=\angle D C E$, the four points $A, D, E$ and $C$ lie on a circle. Thus $60^{\circ}-\gamma=\angle C A E=\angle C D E=\angle P D E=\angle P B E$, whereby the last equality applies due to the chord quadrilateral $D B E P$. In triangle $A B C$, calculate $\angle A B P=$ $180^{\circ}-60^{\circ}-2 \gamma-\left(60^{\circ}-\gamma\right)=60^{\circ}-\gamma$. Thus $B P$ is the bisector of $\angle A B C$. It follows that $P$ is the center of the inscribed circle of triangle $A B C$ and therefore $\angle B A P=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_28_6f8e6b6b48c0be53447cg-01.jpg?height=1125&width=1745&top_left_y=1245&top_left_x=310)

## Marking Scheme:

- $+2 D$ lies on angle bisectors.
- $+2 B P$ is an angle bisector.
- $+1 P$ is the intersection of angle bisectors.
- +2 If $P$ lies on the angle bisectors, we are done.
- $-130^{\circ}$ not calculated.
1. let $A B C$ be a triangle with $\angle B A C=60^{\circ}$. Let $E$ be the point on the side $B C$ such that $2 \angle B A E=$ $\angle A C B$. Let $D$ be the second point of intersection of $A B$ and the circumcircle of the triangle $A E C$ and let $P$ be the second point of intersection of $C D$ and the circumcircle of the triangle $D B E$. Calculate the angle $\angle B A P$.