On the circumcircle of triangle $ABC$, point $P$ is chosen, such that the perpendicular drawn from point $P$ to line $AC$ intersects the circle again at a point $Q$, the perpendicular drawn from point $Q$ to line $AB$ intersects the circle again at a point $R$ and the perpendicular drawn from point $R$ to line $BC$ intersects the circle again at the initial point $P$. Let $O$ be the centre of this circle. Prove that $\angle POC = 90^o$.