An isosceles trapezoid $ABCD$ is circumscribed around a circle. The lateral side $AB$ touches the circle at point $M$, and the base $AD$ at point $N$. Segments $MN$ and $AC$ intersect at point $P$, and $NP: PM=2$. Find

the ratio $AD: BC$.
Draw a line through point $M$ parallel to the bases of the trapezoid, and consider the similar triangles obtained.

## Solution

Let $K$ be the point of tangency of the circle with the base $B C$. Draw a line through point $M$ parallel to $A D$, intersecting the diagonal $A C$ at point $Q$. Denote $A D=4 a, B C=4 b$. Then $A M=A N=2 a, B M=B K=2 b$. From the similarity of triangles $M Q P$ and $N A P$, we find that $M Q=1 / 2 A N=a$, and from the similarity of triangles $A M Q$ and $A B C$, we have ${ }^{A M} /_{A B}=M Q /{ }_{B C}$, or ${ }^{a} / a+b={ }^{a} / 4 b$.

Therefore, $a=3 b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-04.jpg?height=398&width=640&top_left_y=1&top_left_x=709)

## Answer

$3: 1$.