Shnol D....

Given a triangle $ABC$ and an excircle with center $O$, touching side $BC$ and the extensions of sides $AB$ and $AC$. Point $O_{1}$ is symmetric to point $O$ with respect to line $BC$. Find the measure of angle $A$, if it is known that point $O_{1}$ lies on the circumcircle of triangle $ABC$.
From the condition, it follows that $\angle B O C=\angle B O_{1} C=\angle A$. On the other hand, $\angle B O C=180^{\circ}-1 / 2\left(180^{\circ}-\angle B\right)-1 / 2\left(180^{\circ}-\right.$ $\angle C)=1 / 2\left(180^{\circ}-\angle A\right)$. From this, we obtain that $\angle A=60^{\circ}$.