The height of the isosceles trapezoid $ABCD$ with bases $AD$ and $BC$ is $4 \sqrt{3}$, the diagonals of the trapezoid intersect at point $O, \angle AOD=120^{\circ}$. Find the midline of the trapezoid.
The diagonals of an isosceles trapezoid form equal angles with the base, therefore

$$
\angle O A D=\angle O D A=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}
$$

Let $H$ be the foot of the perpendicular dropped from vertex $C$ to $A D$. Then $A H=\frac{1}{2}(A D+B C)$, i.e., segment $A H$ is equal to the midline of the trapezoid. From the right triangle $A C H$, we find that

$$
A H=C H \operatorname{ctg} \angle C A H=C H \operatorname{ctg} 30^{\circ}=4 \sqrt{3} \cdot \sqrt{3}=12
$$

Therefore, the midline of trapezoid $A B C D$ is 12.

## Answer

12.00