3. One of the angles of the trapezoid is $60^{\circ}$. Find the ratio of its bases, if it is known that a circle can be inscribed in this trapezoid and a circle can be circumscribed around this trapezoid.
Answer. $1: 3$

## Solution.

A

![](https://cdn.mathpix.com/cropped/2024_05_06_3f282128bd4470672e40g-1.jpg?height=425&width=626&top_left_y=1775&top_left_x=338)

Since $ABCD$ is a cyclic quadrilateral, it is isosceles, i.e., $AB = CD$. Since $\angle BAD = 60^\circ$, then $\angle ABC = 120^\circ$. The center of the inscribed circle lies at point O, the intersection of the angle bisectors BK and AO of angles BAD and ABC. Since $\angle ABK = 60^\circ = \angle BAK$, triangle $ABK$ is equilateral, meaning that the bisector $AO$ is also a median in this triangle. The bisector OL of angle $BCD$ also passes through point O. Since $O$ is the midpoint of $BK$, OL is the midline of triangle $ABK$ (it passes through the midpoint of $BK$ and is parallel to $AB$), thus $AL = LK$. Similarly, $LK = KD$. Triangles $BCO$ and $LKO$ are equilateral (each angle is $60^\circ$) and their sides are equal ($BO = OK$), therefore $BC = LK = AL = KD$, i.e., $3BC = AD$.