3. An isosceles trapezoid $A B C D$ with bases $A B$ and $D C$ has a circle inscribed in it with the center at point $O$. Find the area of the trapezoid if $O B=b$ and $O C=c$.
Answer: $2 b c$.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-11.jpg?height=449&width=580&top_left_y=1306&top_left_x=818)

Solution. Let $r$ be the radius of the circle, $K$ be the point of tangency of the circle with side $B C$, and $M N$ be the height of the trapezoid passing through point $O$ (see figure). Note that the triangles $O C N$ and $O C K$, as well as the triangles $O B M$ and $O B K$, are equal by three sides. Then

$$
180^{\circ}=\angle N O K+\angle K O M=2 \angle C O K+2 \angle K O B=2 \angle C O B
$$

from which $\angle C O B=90^{\circ}$. Since the trapezoid is inscribed and isosceles, we get $A B+D C=2 B C$. Therefore,

$$
S_{A B C D}=M N \cdot \frac{1}{2}(A B+D C)=2 r \cdot B C=4 S_{B O C}=2 b c
$$