It suffices to prove that $A_1$ is the circumcenter of $\triangle PAQ$; then, $\angle PA_1Q$ is a central angle and $\angle PAQ$ is an inscribed angle of $(PAQ)$, implying the result. Let $H$ be the foot of the altitude from $A$ to $PQ$. Note that since $BC$ and $PQ$ are antiparallel, $\angle ACB=\angle APQ\implies\angle PAH=\angle CAA_1$, meaning that $AA_1$ and $AH$ are isogonal conjugates. Now, note that for any triangle, the circumcenter and the orthocenter are isogonal conjugates. Therefore, if the circumcenter of $\triangle PAQ$ lies on $AA_1$, then the orthocenter of $\triangle PAQ$ must lie on $AH$, which is obviously true. This implies that the circumcenter of $\triangle PAQ$ does indeed lie on $AA_1$. Now consider the intersection of the perpendicular bisector of $PQ$ with the line $AA_1$. Since $\triangle ABC$, and therefore $\triangle PAQ$, is not isosceles, these two lines are not the same. Therefore, if they do intersect, then they intersect in exactly one point. However, it is obvious that $A_1$ lies on this perpendicular bisector of $PQ$. Thus, the two lines must intersect at $A_1$. Since the circumcenter of $\triangle PAQ$ must lie on both $AA_1$ and this perpendicular bisector, $A_1$ is the circumcenter of $\triangle PAQ$ and we are done. $\blacksquare$ EDIT: This should be correct, if a bit wordy.

Let points $H$ and $O$ are orthocenter and circumcenter of triangle $ABC$. Because triangles $APQ$ and $ABC$ are similar then point $A_{1}$ in triangle $APQ$ is the same as intersection of line $AO$ with perpedicular bisector of $BC$ in triangle $ABC$ ($H$ and $O$ are isogonal conjugate), but this is point $O$, so $A_{1}$ is circumcenter of triangle $APQ$.

As said below its enough to show, that $A_1$ is circumcenter of $\Delta APQ$. $\angle A_1AQ = \dfrac{\pi}{2} - \angle ACB = \dfrac{\pi}{2} - \angle APQ$ so circumcenter of $\Delta APQ$ lies on $AA_1$. But it also lies on perp. bisector of $PQ$. As we have $AB \neq AC \implies AP \neq AQ$ this two lines does not coincide, but they have common point $A_1$, so it is the circumcenter.

Fever wrote: As said below its enough to show, that $A_1$ is circumcenter of $\Delta APQ$. $\angle A_1AQ = \dfrac{\pi}{2} - \angle ACB = \dfrac{\pi}{2} - \angle APQ$ so circumcenter of $\Delta APQ$ lies on $AA_1$. But it also lies on perp. bisector of $PQ$. As we have $AB \neq AC \implies AP \neq AQ$ this two lines does not coincide, but they have common point $A_1$, so it is the circumcenter. GOOD SOLUTION