4. The bases of trapezoid $A B C D$ are related by the ratio $A D=4 \cdot B C$, and the sum of angles $\angle A+\angle D=120^{0}$. Points $M$ and $N$ are chosen on the lateral sides such that $C N: N D=B M: M A=1: 2 . \quad$ Perpendiculars drawn from points $M$ and $N$ to the lateral sides of the trapezoid intersect at point 0. Find $A D$, if $A O=1$.
# Solution:

Extend the lateral sides until they intersect at point $S$. The angle $S$

![](https://cdn.mathpix.com/cropped/2024_05_06_b89800d9c16e2aa59f61g-1.jpg?height=271&width=531&top_left_y=1843&top_left_x=1430)
is $60^{\circ}$. Triangles $A S D$ and $B S C$ are similar; the similarity ratio is 4. Moreover, by the condition, $A M$ is twice as long as $B M$. From this, it is not difficult 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 circumcircle of triangle $A S D$, and $A O$ is its radius. We can write the formula for the circumradius $A O = \frac{A D}{2 \cdot \sin \angle S}$. From this, $A D = \sqrt{3}$.

Answer: $\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b89800d9c16e2aa59f61g-1.jpg?height=351&width=556&top_left_y=2211&top_left_x=1389)