Iranian Mathematical Olympiad

Let $T$ be the point of intersection of the perpendicular bisector of $BC$ and circle $\omega$, so $TD$ is a diagonal of $\omega$ and $\angle DFT=90^{\circ }$. Since $D$ is the midpoint of arc $\text{widearc}BAC$, $FD$ is the angle bisector of $\angle BFC$. Therefore $(JEBC)=-1$ where $J$ is the intersection point of line $TF$ and the extension of $BC$.



$$\angle EJF=\frac{1}{2}(\text{widearc}CT-\text{widearc}BF)=\frac{1}{2}(\text{widearc}BT-\text{widearc}BF)=\frac{1}{2}\text{widearc}FT=\angle TAF=\angle EPF$$ So quadrilateral $PEFJ$ is cyclic and consequently $\angle JPE=90^{\circ }$. This fact and $(JEBC)=-1$ implies that $PE$ is the angle bisector of $\angle BPC$ and this is what we wanted to prove.