Since $AB \cdot AQ = AP^2 = AC \cdot AR$, quadrilateral $BCRQ$ is cyclic. Let $O$ be the center of circle ($BCRQ$). Denote by $O_1$ and $O_2$ the centers of circles ($BPC$) and ($QPR$). We will show that lines $BO_1$, $CO_1$, $QO_2$, $RO_2$ are equidistant from $O$. Since $OB = OC = OQ = OR$, it suffices to establish the equality of (directed) angles $\angle OCO_1 = \angle O_1BO = \angle OQO_2 = \angle O_2RO$. Here the first and last equalities are obvious from symmetry about the perpendicular bisectors of $BC$ and $QR$.

Рис. 5

It remains to prove the equality $\angle O_1BO = \angle OQO_2$ (*). By angle chasing we obtain $\angle OQO_2 = \angle OQR - \angle O_2QR = (90^\circ - \angle RCQ) - (90^\circ - \angle RPQ) = \angle RPQ - \angle RCQ$. Similarly $\angle O_1BO = \angle BPC - \angle BQC$. Thus, (*) is equivalent to the equality $\angle RPQ - \angle RCQ = \angle BPC - \angle BQC$ or $\angle BQC - \angle RCQ = \angle BPC - \angle RPQ$ (**). From the tangency of circles ($BPQ$) and ($CPR$) it follows that $\angle RPQ = \angle RCP + \angle PBQ$, which equals (from the sum of angles in quadrilateral $BPCA$) $\angle BPC - \angle BAC$. Therefore, (**), transforms into $\angle BQC - \angle RCQ = \angle BAC$, which holds true. The problem is solved.