In the plane of a triangle
in its exterior
we draw the triangles
so that
Prove that
a.)
and
b.)
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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.) $QR = RP.$
Školjka