Solution. Here we use directed angles measured in the counterclockwise direction.
Since $AC_1$ is the common chord of the circles $O$ and $O_2$, we see that $OO_2 \perp AC_1$. Therefore, $OO_2 \perp AO_1$ is equivalent to that $O_1$ lies on $AC_1$.
Assume that $O_1$ lies on $AC_1$. (Claim that $O_2$ lies on $AC$.)
Since $O$ and $O_1$ are circumcenters of $ABC$ and $AB_1C$, it is easy to see that
$$\angle O_1C_1O_1 = \angle O_1CA = \angle C_1AO = \angle C_1AC + \angle CAO = \angle AC_1O_1 + \angle OCA = \angle OCO_1.$$
Therefore, $O, O_1, C_1, C$ are cyclic.
Consider $\triangle AB_2C_1$ with circumcenter $O_2$. Since $OA = OC_1$ and $O_2A = O_2C_1$, $\triangle AOO_2 \cong \triangle C_1OO_2$ and $\angle C_1O_2O = \frac{1}{2}\angle C_1O_2A = \angle C_1B_2A$.
$$\begin{aligned} \angle B_2AC_1 &= \frac{1}{2}(\pi - \angle A_1B_1) = \frac{\pi}{2} - \angle ACB_1 = \frac{\pi}{2} - \angle ACB - \angle BCB_1 \\ &= \frac{\pi}{2} - \angle ACB - \angle BAC = \angle CBA - \frac{\pi}{2}. \end{aligned}$$

$$\begin{aligned} \angle AC_1B_2 &= \angle BC_1B_2 + \angle AC_1B = \angle B_2AC_1 + \angle ACB = \angle CBA + \angle ACB - \frac{\pi}{2} \\ &= \frac{\pi}{2} - \angle BAC. \end{aligned}$$
$$$$
$\therefore \angle$ C_1B_2A = $\pi - \angle$ B_2AC_1 - $\angle$ AC_1B_2 = $\pi + \angle$ BAC - $\angle$ CBA.
$$$$
$$\begin{aligned} \angle C_1O_1O &= \angle CO_1O + \angle C_1O_1C = \angle CC_1O + \angle C_1OC = \frac{1}{2}(\pi - \angle C_1OC) + \angle C_1OC \\ &= \frac{\pi}{2} + \angle C_1AC = \frac{\pi}{2} + \angle BAC - \angle B_2AC_1 \\ &= \pi + \angle BAC - \angle CBA. \end{aligned}$$
$$$$
Thus, $\angle C_1O_2O = C_1O_1O$, and then, $O, O_2, O_1, C_1, C$ lie on the same circle.
Let $O'_2$ be the intersection point of AC and the perpendicular bisector of AC. Then $\angle O'_2OO_1 = \angle O_1AO'_2 = \angle O'_2CO_1$. Therefore, $O'_2$ also lies on the circle passing through $O, O_2, O_1, C_1, C$. Since $O, O_2, O'_2$ lie on the perpendicular bisector of AC, where $O_2$ and $O'_2$ are different from O (since $\angle CBA > 90^\circ$, O is outside $\triangle ABC$), this implies that $O_2 = O'_2$. (Because a line can intersect a circle in at most 2 points.) Therefore, $O_2$ lies on AC.