10.233. Find the midline of an isosceles trapezoid with height $h$, if the lateral side is seen from the center of the circumscribed circle at an angle of $120^{\circ}$.
## Solution.

In trapezoid $ABCD$ (Fig. 10.41) $BC \| AD, AB=CD, O$ is the center of the circumscribed circle, $\angle AOB=120^{\circ}$. Then the inscribed $\angle ADB=\frac{1}{2} \angle AOB=60^{\circ}$. $BE$ is the height of the trapezoid, $BE=h$. Since the trapezoid is isosceles, the length

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0737.jpg?height=543&width=486&top_left_y=19&top_left_x=193)

Fig. 10.42

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0737.jpg?height=586&width=520&top_left_y=12&top_left_x=654)

Fig. 10.43

$ED$ is equal to the length of the midline of the trapezoid. From $\triangle BED\left(\angle BED=90^{\circ}\right)$: $DE=BE \operatorname{ctg} \angle ADB=\frac{h \sqrt{3}}{3}$.

Answer: $\frac{h \sqrt{3}}{3}$.