11.3. 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.
# Solution.

Since ABCD is a cyclic quadrilateral, it is isosceles, i.e., $\mathrm{AB}=\mathrm{CD}$.

Since the angle $\mathrm{BAD}=60^{\circ}$, then $\mathrm{ABC}=120^{\circ}$.

The center of the inscribed circle lies at point O, the intersection of the angle bisectors BK and AO of angles BAD and

$\mathrm{ABC}$.

Since angles $\mathrm{ABK}=60^{\circ}=\mathrm{BAK}$, triangle $\mathrm{ABK}$ is equilateral, meaning bisector AO

is the 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 $\mathrm{ABK}$ (passes through the midpoint of $\mathrm{BK}$ and is parallel to $\mathrm{AB}$), hence $\mathrm{AL}=\mathrm{LK}$. Similarly, $\mathrm{LK}=\mathrm{KD}$. Triangles $\mathrm{BCO}$ and $\mathrm{LKO}$ are equilateral (angles are $60^{\circ}$) and their sides are equal ( $\mathrm{BO}=\mathrm{OK}$ ), thus $\mathrm{BC}=\mathrm{LK}=\mathrm{AL}=\mathrm{KD}$, i.e., $3 \mathrm{BC}=\mathrm{AD}$.

A

![](https://cdn.mathpix.com/cropped/2024_05_06_5eca02a57588bb860b00g-2.jpg?height=260&width=420&top_left_y=2017&top_left_x=361)

Answer. 1:3