Task 2. (Option 2)

The bisector $D E$ of triangle $A D C$ is extended to intersect the circumscribed circle at point $B$. It is known that the length $B D=1$ and the value $\sin \angle A D C=a$. Find the area of quadrilateral $A B C D$.
# Solution:

For the sought area, we have (see fig.):

$S=\frac{1}{2} B D \cdot A C \cdot \sin \angle D E C=\frac{1}{2} B D \cdot 2 R \cdot \sin \angle D E C$

From $\triangle A D E$ and $\triangle A B E$, it is clear that $\angle D E C=\alpha+\frac{\beta}{2}=\angle D A B$,

![](https://cdn.mathpix.com/cropped/2024_05_06_96a99fe03936e779e014g-03.jpg?height=525&width=514&top_left_y=366&top_left_x=1362)

where $\alpha=\angle D A E, \frac{\beta}{2}=\angle A D E=\angle C D E=\angle B A E=\frac{1}{2} \cup B C$. Therefore:

$$
\begin{aligned}
& S=\frac{1}{2} B D \cdot(2 R \cdot \sin \angle D E C) \sin \angle A D C=\frac{1}{2} B D \cdot(2 R \cdot \sin \angle D A B) \sin \angle A D C= \\
& =\frac{1}{2} B D \cdot B D \cdot \sin \angle A D C=\frac{1}{2} B D^{2} \sin \angle A D C=\frac{1}{2} l^{2} \sin \angle A D C
\end{aligned}
$$

Answer: $\frac{1}{2} l^{2} \sin \angle A D C$