A trapezoid with height $h$ is inscribed in a circle. The lateral side of the trapezoid is seen from the center of the circle at an angle of $120^{\circ}$. Find the midline of the trapezoid.
Draw a line through the vertex of the smaller base of the trapezoid, parallel to one of the diagonals, or use the fact that the projection of the diagonal of an isosceles trapezoid onto the larger base is equal to the midline of the trapezoid.

## Solution

## First Method.

Let $O$ be the center of the circle circumscribed around trapezoid $ABCD$ with bases $AD > BC$. Since the trapezoid is isosceles,

$$
AC = BD, \angle CAD = \angle BDA = \frac{1}{2} \angle BOA = 60^\circ.
$$

Draw a line through vertex $C$ parallel to $BD$, intersecting the extension of $AD$ at point $K$. Then $\triangle ACK$ is an equilateral triangle,

$$
AK = 2h \cos 60^\circ = \frac{2h}{\sqrt{3}}
$$

Therefore, the midline of trapezoid $ABCD$ is

$$
\frac{AD + BC}{2} = \frac{AD + DK}{2} = \frac{1}{2} AK = \frac{h}{\sqrt{3}}
$$

Second Method.

Let $O$ be the center of the circle circumscribed around trapezoid $ABCD$ with bases $AD > BC$. Since the trapezoid is isosceles,

$$
\angle CAD = \angle BDA = \frac{1}{2} \angle BOA = 60^\circ
$$

Let $CK$ be the height of the trapezoid, then

$$
AK = h \cot 60^\circ = \frac{h}{\sqrt{3}}
$$

but $AK$ is equal to the midline of trapezoid $ABCD$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-02.jpg?height=503&width=698&top_left_y=729&top_left_x=680)

Answer

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