International Mathematical Olympiad 1975 Problem 3

On the sides of an arbitrary triangle $ABC$, triangles $ABR, BCP, CAQ$ are constructed externally with $\angle CBP = \angle CAQ = 45°$, $\angle BCP = \angle ACQ = 30°$, $\angle ABR = \angle BAR = 15°$. Prove that $\angle QRP = 90°$ and $QR = RP$.