IMO 1975 Problem 3

In the plane of a triangle $ABC,$ in its exterior $,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$ , $\angle BCP = \angle QCA = 30^{\circ}$ , $\angle ABR = \angle RAB = 15^{\circ}$ .

Prove that a.) $\angle QRP = 90\,^{\circ},$ and b.) $QR = RP.$