[ Ratio of areas of triangles with a common angle ]

On the side $A B$ of trapezoid $A B C D$, a point $M$ is taken such that $A M: B M=2: 3$. On the opposite side $C D$, a point $N$ is taken such that the segment $M N$ divides the trapezoid into parts, one of which is three times larger in area than the other. Find the ratio $C N: D N$, if $B C: A D=1: 2$.

#
Continue the lateral sides of the trapezoid to their intersection at point $P$ and use the equality

$$
S_{\triangle \mathrm{MPN}}=\frac{P M}{P A} \cdot \frac{P N}{P D} \cdot S_{\Delta \mathrm{APD}}
$$

Consider two cases.

## Solution

Let the extensions of the lateral sides $A B$ and $C D$ of the trapezoid intersect at point $P$. Then $B C$ is the midline of triangle $A P D$. Therefore,

$$
\frac{P M}{P A}=\frac{8}{10}=\frac{4}{5}
$$

Let $S$ be the area of trapezoid $A B C D$. Then the area of triangle $B P C$ is $\frac{1}{3} S$.

Assume that the area of quadrilateral $A M N D$ is three times smaller than the area of quadrilateral $M B C N$. Then

$$
S_{\triangle \mathrm{APD}}=\frac{4}{3} S, S_{\mathrm{MBCN}}=\frac{3}{4} S, S_{\Delta \mathrm{MPN}}=\frac{1}{3} S+\frac{3}{4} S=\frac{13}{12} S
$$

and since

$$
S_{\triangle \mathrm{MPN}}=\frac{P M}{P A} \cdot \frac{P N}{P D} S_{\Delta \mathrm{APD}}
$$

we have

$$
\frac{13}{12} S=\frac{4}{5} \cdot \frac{P N}{P D} \cdot \frac{4}{3} S
$$

from which we find that $\frac{P N}{P D}=\frac{65}{64}>1$, i.e., $P N>P D$, which is impossible since point $N$ must belong to segment $C D$.

Now let the area of quadrilateral $M B C N$ be three times smaller than the area of quadrilateral $A M N D$. Then

$$
S_{\triangle \mathrm{APD}}=\frac{4}{3} S, S_{\mathrm{MBCN}}=\frac{1}{4} S, S_{\Delta \mathrm{MPN}}=\frac{1}{3} S+\frac{1}{4} S=\frac{7}{12} S
$$

and since

$$
S_{\Delta \mathrm{MPN}}=\frac{P M}{P A} \cdot \frac{P N}{P D} S_{\Delta \mathrm{APD}}
$$

we have

$$
\frac{7}{12} S=\frac{4}{5} \cdot \frac{P N}{P D} \cdot \frac{4}{3} S
$$

from which we find that $\frac{P N}{P D}=\frac{35}{64}$. Therefore,

$$
\frac{P N}{N D}=\frac{35}{29}, \frac{C N}{N D}=\frac{35-32}{29}=\frac{3}{29}
$$

## Answer

$3: 29$.