MathNet 1993 — Part 01 Question 10

$ABCD$ is a convex quadrilateral with $\angle BAC = 30^\circ$, $\angle CAD = 20^\circ$, $\angle ABD = 50^\circ$, $\angle DBC = 30^\circ$. If the diagonals intersect at $P$, show that $PC = PD$.