[Area of a Quadrilateral]

Find the area of quadrilateral $ABCD$, if $AB=BC=3\sqrt{3}$, $AD=DC=\sqrt{13}$, and vertex $D$ lies on a circle of radius 2 inscribed in angle $ABC$, with $\angle ABC=60^{\circ}$.
Since $A B=B C$ and $A D=D C$, points $B$ and $D$ lie on the perpendicular bisector of segment $A C$. Therefore, $B D$ is the angle bisector of $\angle A B C$. This means the center $O$ of the given circle lies on the ray $B D$. Since $\angle A B C=60^{\circ}$, triangle $A B C$ is equilateral. Let points $B$ and $D$ lie on opposite sides of line $A C$ (Fig.1), and let $E$ be the intersection of $B D$ and $A C$. Since $E$ is the midpoint of $A C$,

\[
\begin{gathered}
D E=\sqrt{A D^{2}-A E^{2}}=\sqrt{(\sqrt{13})^{2}-\left(\frac{3 \sqrt{3}}{2}\right)^{2}}=\sqrt{13-\frac{27}{4}}=\frac{5}{2} \\
B E=A B \cdot \cos 30^{\circ}=3 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=\frac{9}{2}
\end{gathered}
\]

Let $M$ be the point of tangency of the circle with line $A B$. Since point $E$ lies on segment $B D$, $B D=B E+E D=\frac{\frac{9}{2}}{2}+$ $\frac{5}{2}=7$. At the same time, $B O=2 \cdot O M=4$ and $O D=2$, so $B D=B O+O D=6$ (if point $O$ lies on segment $B D$), or $B D=B O-O D=4-2=2$ (if point $O$ lies on the extension of segment $B D$ beyond point $D$), which is impossible. The only remaining case is when points $B$ and $D$ lie on the same side of line $A C$ (Fig.2). Then

\[
B D=B E-D E=\frac{9}{2}-\frac{5}{2}=2
\]

Therefore,

\[
S_{A B C D}=\frac{1}{2} \cdot B D \cdot A C=\frac{1}{2} \cdot 2 \cdot 3 \sqrt{3}=3 \sqrt{3} .
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_4541620ef230b5fef331g-01.jpg?height=906&width=874&top_left_y=1582&top_left_x=592)

Problem