In triangle $A B C$, angle $C$ is $60^{\circ}$, and the radius of the circle circumscribed around this triangle is $2 \sqrt{3}$.

On side $A B$, a point $D$ is taken such that $A D=2 D B$ and $C D=2 \sqrt{2}$. Find the area of triangle $A B C$.

#
Using the theorem of intersecting chords, find $O D$ ($O$ - the center of the circle). Triangle $O D C$ is a right triangle.

## Solution

Let $O$ be the center of the circle circumscribed around triangle $A B C$, with $R=2 \sqrt{3}$ as its radius. Then,

$$
A B=2 R \sin \angle C=4 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=6
$$

Therefore, $A D=4, B D=2$.

Since $(R-O D)(R+O D)=A D \cdot D B$, we have

$$
O D^{2}=R^{2}-A D \cdot D B=12-8=4
$$

Therefore, $O D=2$.

Since $O D^{2}+D C^{2}=O C^{2}$, triangle $O D C$ is a right triangle $\left(\angle O D C=90^{\circ}\right)$. Since triangle $O D B$ is isosceles, we have

$$
\angle D O B=\angle O B D=30^{\circ} .
$$

Therefore, $\angle C D B=30^{\circ}$.

Let $C M$ be the altitude of triangle $A B C$. Then $C M=\frac{1}{2} C D=\sqrt{2}$. Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot C M=\frac{6 \sqrt{2}}{2}=3 \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-10.jpg?height=420&width=548&top_left_y=585&top_left_x=755)

Answer

$3 \sqrt{2}$.