Mathematica competitions in Croatia

The triangle $OCB$ is isosceles right triangle because $\overline{OB}$ and $\overline{OC}$ are both radii of the circle with the centre $O$. Hence $\angle CBA=\angle CBO=45^{\circ }$.



Inscribed angles $\angle CPA$ and $\angle CBA$ over the chord $\overline{CA}$ are equal, so we have $$\angle QPR=\angle CPA=\angle CBA=45^{\circ }.$$

By Thales' Theorem the angle $\angle APB$ is right, so the angle $\angle BPR$ must be right as well. The quadrilateral $BQRP$ is cyclic because it has two opposite right angles ($\angle RQB$ and $\angle BPR$), so the inscribed angles over the chord $QR$ are equal, which gives $\angle QBR=\angle QPR=45^{\circ }$. Hence $\angle BRQ=180^{\circ }-\angle RQB-\angle QBR=45^{\circ }$, so $\angle BRQ=45^{\circ }=\angle QBR$ and we get $|BQ|=|QR|$.