EGMO 2020 Problem 5

Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$ . The circumcircle $\Gamma$ of $ABC$ has radius $R$ . There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$ . The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$ .

Prove $P$ is the incentre of triangle $CDE$ .