Trapezoids $A B C D$ and $A C D E$ with equal larger bases $A D$ and $A C$ respectively are inscribed in a circle. What is the radius of this circle if the area of triangle $A D E$ is $1+\sqrt{3}$, and the angle $C O D$ is $60^{\circ}$, where $O$ is the point of intersection of the diagonals of trapezoid $A B C D$?
$\angle C O D=\frac{U C D+U A B}{2}$.

## Solution

Arcs enclosed between parallel chords are equal. Therefore, $\cup A B=\cup C D=\cup A E$. Since chords $A C$ and $A D$ are equal, then

$$
\cup A B+\cup B C=\cup A E+\cup E D
$$

Thus, $\cup B C=\cup E D$.

Let $\cup A B=\alpha, \cup B C=\beta$. Then

$$
\angle C O D=\frac{\cup C D+\cup A B}{2}=\alpha=60^{\circ}, 2 \beta=360^{\circ}-3 \alpha=180^{\circ}
$$

Therefore, $\beta=90^{\circ}$. Hence,

$$
\angle A D E=\frac{1}{2} \cup A E=\frac{\alpha}{2}=30^{\circ}, \angle D A E=\frac{1}{2} \cup E D=\frac{\beta}{2}=45^{\circ}
$$

Let $R$ be the desired radius. Then

$$
S_{\triangle \mathrm{ADE}}=\frac{1}{2} A E \cdot D E \sin \angle A E D=\frac{1}{2} \cdot 2 R \sin 30^{\circ} \cdot 2 R \sin 45^{\circ} \cdot \sin 75^{\circ}=\frac{R^{2}(\sqrt{3}+1)}{4}=\sqrt{3}+1
$$

Therefore, $R^{2}=4$.

## Answer

2.