$A B C D$ - is an isosceles trapezoid.

## Solution

Since $-A B=-C D$, then $B C \| A D$. Therefore, $A B C D$ - is an isosceles trapezoid. $\angle B D A=1 / 2 \angle A O B=30^{\circ}$, hence, $K D=B K \operatorname{ctg} 30^{\circ}=6 \sqrt{3}$, On the other hand, $K D=1 / 2(A D+B C)=2 B C$, from which $A K=B C=3 \sqrt{3}, A B^{2}=$ $A K^{2}+B K^{2}=63$. Triangle $A O B$ is equilateral, therefore, $S_{C O D}=S_{A O B}=\frac{\sqrt{3}}{4} A B^{2}=\frac{63 \sqrt{3}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_82cd6cdb4fc29ee56532g-86.jpg?height=280&width=483&top_left_y=1029&top_left_x=793)

## Answer

$\frac{63 \sqrt{3}}{4}$.
$$
\begin{aligned}
& \text { [Isosceles, inscribed, and circumscribed trapezoids] } \\
& \text { [ Right triangle with an angle of \$30^\circ\$ ] } \\
& \text { [ Pythagorean theorem (direct and inverse). ] } \\
& \text { [ Equilateral (equiangular) triangle ] }
\end{aligned}
$$

Quadrilateral $A B C D$ is inscribed in a circle with center $O, \angle B O A=\angle C O D=60^{\circ}$. The perpendicular $B K$, dropped to side $A D$, is equal to $6 ; A D=3 B C$.

Find the area of triangle $C O D$.