JBMO 2002 Problem 1

The triangle $ABC$ has $CA = CB$ . $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$ ). $D$ is the foot of the perpendicular from $C$ to $PB$ . Show that $PA + PB = 2 \cdot PD$ .