In trapezoid $A B C D$, angles $A$ and $D$ at the base $A D$ are respectively $60^{\circ}$ and $30^{\circ}$. Point $N$ lies on the base $B C$, such that $B N: N C=2$. Point $M$ lies on the base $A D$, line $M N$ is perpendicular to the bases of the trapezoid and divides its area in half. Find the ratio $A M: M D$.
Let $N C=a$ and express the height of the trapezoid in terms of $a$.

## Solution

Let $N C=a, N M=h$. Draw lines through point $N$ parallel to $A B$ and $C D$, intersecting the base $A D$ at points $P$ and $Q$ respectively. Then

\[
\begin{gathered}
A P=B N=2 a, Q D=N C=a \\
P M=h \cot 60^{\circ}=\frac{h \sqrt{3}}{3}, Q M=h \cot 30^{\circ}=h \sqrt{3}
\end{gathered}
\]

From the equality of the areas of trapezoids $A B N M$ and $M N C D$, it follows that

\[
B N+A M=N C+M D, \text { or } 2 a+2 a+\frac{h \sqrt{3}}{3}=a+a+h \sqrt{3}.
\]

From this, we find that $h=a \sqrt{3}$. Therefore,

\[
A M=2 a+a=3 a, M D=3 a+a=4 a, A M: M D=3: 4
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-31.jpg?height=343&width=824&top_left_y=1375&top_left_x=617)

## Answer

$3: 4$.