In an isosceles triangle $ABC (AB = BC)$, the altitude $CD$ is drawn. The angle $BAC$ is equal to $\alpha$. The radius of the circle passing through points $A, C$, and $D$ is $R$. Find the area of triangle $ABC$.

#
From point $D$, the segment $A C$ is seen at a right angle, which means this point lies on the circle with diameter $A C$. Since a unique circle passes through three points not lying on the same line, the circle with diameter $A C$ is the circle mentioned in the problem. Let $O$ be its center. Then $O$ is the midpoint of the base $A C$ of the isosceles triangle $A B C$, so $B O$ is the height of this triangle. From the right triangle $O A B$, we find that $B O = O A \operatorname{tg} \angle O A B = R \operatorname{tg} \alpha$. Therefore,

$$
S_{\triangle A B C} = \frac{1}{2} A C \cdot B O = \frac{1}{2} \cdot 2 R \cdot R \operatorname{tg} \alpha = R^2 \operatorname{tg} \alpha \text{.}
$$

## Answer

$R^2 \operatorname{tg} \alpha$.