On the plane, there is an angle of $60^{\circ}$. A circle touches one side of this angle, intersects the other side at points $A$ and $B$, and intersects the angle bisector at points $C$ and $D$. $A B = C D = \sqrt{6}$. Find the area of the circle bounded by this circle.
Let $K$ be the vertex of the angle, $O$ be the center of the circle, $R$ be its radius, $M$ be the point of tangency with the side of the angle, and $L$ be the midpoint of the chord $CD$.

From the condition, it is clear that $KO$ is the bisector of the angle $AOC$, so $\angle OKC = 15^\circ$. Therefore, $\angle OKM = 45^\circ$, $OM = R$, $OK = R\sqrt{2}$, $\left(R\sqrt{2} \sin 15^\circ\right)^2 = OL^2 = R^2 - \frac{3}{2}$, or $R^2\left(1 - \cos 30^\circ\right) = R^2 - \frac{3}{2}$, from which $R^2 = \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} = \sqrt{3}$.

## Answer

$\pi \sqrt{3}$.