A circle is inscribed in a convex quadrilateral $A B C D$ with its center at point $O$, and $A O=O C, B C=5, C D=$ 12, and angle $D A B$ is a right angle.

Find the area of quadrilateral $A B C D$.
Prove the equality of triangles $D A B$ and $D C B$.

## Solution

Let $M$ and $N$ be the points of tangency of the inscribed circle with sides $A B$ and $B C$ respectively. Triangles $A M O$ and $C N O$ are equal by the leg and hypotenuse. Therefore, $A M=C N$, and since $B M=B N$, then $A B=B C=5$. Similarly, $A D=D C=12$. Therefore, triangles $D A B$ and $D C B$ are equal by three sides. Hence, $S_{A B C D}=2 S_{D A B}=60$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-17.jpg?height=332&width=600&top_left_y=586&top_left_x=729)

## Answer

60.