Question 3
On the sides of an arbitrary triangle $ABC$, triangles $BPC$, $CQA$, and $ARB$ are externally erected such that
$$\angle PBC=\angle CAQ=45^\circ,$$
$$\angle BCP=\angle QCA=30^\circ,$$
$$\angle ABR=\angle BAR=15^\circ.$$
Prove that $\angle QRP=90^\circ$ and $QR=RP$.
$$\angle PBC=\angle CAQ=45^\circ,$$
$$\angle BCP=\angle QCA=30^\circ,$$
$$\angle ABR=\angle BAR=15^\circ.$$
Prove that $\angle QRP=90^\circ$ and $QR=RP$.