Problem 1491

1. **Reflecting the Center $O$ Across One Side of the Angle:**
Let $O$ be the center of the inscribed circle, and let $A$ be the reflection of $O$ across one side of the angle. This means that $A$ is equidistant from the side of the angle as $O$ but on the opposite side.

2. **Tangents from $A$ to the Circle:**
Tangents to the circle from $A$ intersect the other side of the angle at points $B$ and $C$. Since $A$ is the reflection of $O$, the tangents from $A$ to the circle are equal in length. Therefore, $AB = AC$.

3. **Circumcircle of $\triangle ABC$:**
Let $(ABC)$ be the circumcircle of $\triangle ABC$. We need to show that the circumcenter of $\triangle ABC$ lies on the bisector of the original angle.

4. **Intersection Points $Q$ and $R$:**
Let $(ABC)$ intersect the side of the angle $O$ was reflected across at points $Q$ and $R$. Because $QR$ bisects $AO$, it follows from the incenter-excenter lemma that $Q$ and $R$ are the midpoints of the arcs $AB$ not including $C$ and arc $AC$ not including $B$, or arc $AB$ including $C$ and arc $AC$ including $B$ (not necessarily in that order depending on the configuration).

5. Isosceles Trapezoid Formation:
Points $Q, B, C, R$ form an isosceles trapezoid. This is because $Q$ and $R$ are the midpoints of the arcs, and thus $QB = QC$ and $RB = RC$. The line $TO$ (where $T$ is the point where the angle bisector intersects the side of the angle) coincides with the perpendicular bisector of the bases of this trapezoid.

6. Conclusion:
Since the perpendicular bisector of the bases of the isosceles trapezoid $Q, B, C, R$ is the angle bisector, the circumcenter of $\triangle ABC$ lies on the bisector of the original angle.

$\blacksquare$