4. The bases of trapezoid $A B C D$ are related by $A D=4 \cdot B C$, and the sum of angles $\angle A+\angle D=120^{\circ}$. Points $M$ and $N$ are chosen on the lateral sides such that $C N: N D=B M: M A=1: 2$. Perpendiculars drawn from points $M$ and $N$ to the lateral sides of the trapezoid intersect at point $O$. Find $A D$, if $A O=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_09141e68ae4e2c0781beg-1.jpg?height=277&width=539&top_left_y=1797&top_left_x=1409)
# Solution:

Extend the lateral sides until they intersect at point $S$. Angle $S$ is $60^{\circ}$. Triangles $A S D$ and $B S C$ are similar; the similarity coefficient is 4. Moreover, by the condition, $A M$ is twice as long as $B M$. From this, it is easy to notice that $B S = B M$. Therefore, $A M = M S$. Thus, $M O$ is the perpendicular bisector of side $A S$. Similarly, $N O$ is the perpendicular bisector of $D S$. Therefore, point $O$ is the center of the circumscribed circle around triangle $A S D$, and $A O$ is its radius. We can write the formula for the radius of the circumscribed circle $A O = \frac{A D}{2 \cdot \sin \angle 5}$. From this, $A D = \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_09141e68ae4e2c0781beg-1.jpg?height=383&width=591&top_left_y=2212&top_left_x=1360)

Answer: $\sqrt{3}$.