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Chord $AB$ subtends an arc of a circle equal to $120^{\circ}$. Point $C$ lies on this arc, and point $D$ lies on chord $AB$. Given that $AD=2, BD=1, DC=\sqrt{2}$.

Find the area of triangle $ABC$.

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Let $O$ be the center of the circle. Then $A O=\mathrm{AB} / 2 \sin 60^{\circ}=\sqrt{3}$.

Since $A D=2, A O=\sqrt{3}$, and $\angle D A O=30^{\circ}$, triangle $A O D$ is a right triangle, and $O D=1$.

Therefore, triangle $O D C$ is a right triangle

$\left(O C^{2}=O D^{2}+C D^{2}\right)$. Thus, $\angle C D A=\angle C D O-\angle A D O=90^{\circ}-2 \angle D B O=30^{\circ}$.

Therefore, $S_{A B C}=1 / 2 A B \cdot A D \sin 30^{\circ}=\frac{3 \sqrt{2}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-14.jpg?height=649&width=649&top_left_y=1716&top_left_x=710)

## Answer

$\frac{3 \sqrt{2}}{4}$.

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