Question 6
Consider the triangle $ABC$ with $\angle BCA \gt 90^{\circ}$. 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 that $P$ is the incentre of triangle $CDE$.
Prove that $P$ is the incentre of triangle $CDE$.