Selection tests for the Gulf Mathematical Olympiad 2013

Assume $AP=AQ$. Because $OP=OQ$, the line $AO$ is perpendicular to $PQ$. But since $\angle C_{1}AO=90^{\circ }-\angle ACB$ then $\angle B_1{C}_{1}A=\angle ACB$, and therefore, quadrilateral $BC{B}_1{C}_1$ is cyclic.

Using the power of the point $O$ with respect to the circumcircle of $BC{B}_1{C}_1$ we get $$O{B}_1=\frac{O{B}_1\cdot OB}{R}=\frac{O{C}_1\cdot OC}{R}=O{C}_1,$$ where $R$ is the circumradius of triangle $ABC$. Therefore, $B{B}_1=C{C}_1$, that is the minors $\widehat{B{B}_1}$ and $\widehat{C{C}_1}$ of the circumcircle of $BC{B}_1{C}_1$ have the same length. This is equivalent to saying that $\angle CB{C}_1=\angle B_{1}CB$, and therefore $AB=AC$.