Iveliev B.M.

Let $ABC$ be an acute-angled triangle, $CC_1$ its bisector, $O$ the center of the circumscribed circle. The intersection point of the line $OC_1$ with the perpendicular dropped from vertex $C$ to side $AB$ lies on the circumscribed circle $\Omega$ of triangle $AOB$. Find the angle $C$.
Let $D$ be the point of intersection of $OC_{1}$ with the given perpendicular (see figure). Since $D$ lies on $\Omega$ and $AO=OB$, then $\angle ADC_{1}=\angle BDC_{1}$. Therefore,

$AD:BD = AC_{1}:BC_{1} = AC:BC$. On the other hand, since $CD \perp AB$, then $AC^2 + BD^2 = AD^2 + BC^2$. From these equalities, it follows that $AC = AD$, which means that point $D$ is symmetric to $C$ with respect to $AB$. Hence, the line $CC_{1}$ intersects the perpendicular bisector of $AB$ at a point symmetric to $O$. Since the intersection of the angle bisector and the perpendicular bisector lies on the circumcircle, we obtain that the chord $AB$ bisects the perpendicular radius. Therefore, the angle $C$ subtending this arc is $60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-09.jpg?height=1237&width=969&top_left_y=-1&top_left_x=545)

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