[Isosceles, inscribed and circumscribed trapezoids] [Circle inscribed in an angle $\quad$]

On the diagonal $AC$ of the convex quadrilateral $ABCD$, there is the center of a circle of radius $r$ that is tangent to the sides $AB, AD$, and $BC$. On the diagonal $BD$, there is the center of a circle of the same radius $r$ that is tangent to the sides $BC$, $CD$, and $AD$. Find the area of the quadrilateral $ABCD$, given that the specified circles touch each other externally.
$A B C D$ - is an isosceles trapezoid.

## Solution

Note that $A D \| B C$ (common external tangents to two equal circles). Let $O_{1}$ and $O_{2}$ be the centers of the first and second circles, respectively.

Since $A C$ and $D B$ are the angle bisectors of angles $A$ and $D$ of trapezoid $A B C D$, triangles $A B C$ and $B C D$ are isosceles. Therefore, trapezoid $A B C D$ is isosceles.

Let $B K$ be the height of the trapezoid. Then

$$
2 r=O_{1} O_{2}=\frac{1}{2}|A D-B C|=A K=B K
$$

Therefore, $\angle B A D=\angle C D A=45^{\circ}$. Consequently,

$$
\begin{gathered}
B C=A B=B K \sqrt{2}=2 r \sqrt{2}, A D=B C+2 A K=2 r \sqrt{2}+4 r \\
S_{\mathrm{ABCD}}=\frac{1}{2}(A D+B C) B K=4 r^{2}(1+\sqrt{2})
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-05.jpg?height=398&width=1129&top_left_y=1&top_left_x=470)

## Answer

$4(\sqrt{2}+1) r^{2}$.