Prokopenko D:

Around triangle $A B C$, a circle $\Omega$ is described. Let $L$ and $W$ be the points of intersection of the angle bisector of angle $A$ with side $B C$ and circle $\Omega$, respectively. Point $O$ is the center of the circumcircle of triangle $A C L$. Restore triangle $A B C$ if the circle $\Omega$ and points $W$ and $O$ are given.
Let $O^{\prime}$ be the center of $\Omega$. Then the lines $O^{\prime} O$ and $O^{\prime} W$ are perpendicular to the sides $A C$ and $B C$, so the directions of these sides are known. Moreover,

$\angle C O L=2 \angle C A L=2 \angle L C W$ and, therefore, $\angle O C W=90^{\circ}$ (see the figure). Consequently, $C$ is the intersection point of the circle $\Omega$ and the circle with diameter $O W$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-12.jpg?height=929&width=896&top_left_y=-4&top_left_x=581)

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