In trapezoid $A B C D$, base $A B$ is three times the length of base $C D$. On base $C D$, a point $M$ is taken such that $M C$ $=2 M D$. $N$ is the point of intersection of lines $B M$ and $A C$. Find the ratio of the area of triangle $M N C$ to the area of the entire trapezoid.
$S_{A D C}=1 / 3 S_{A B C}=1 / 4 S_{A B C D}$, since $A B=3 C D$.

From the similarity of triangles $M N C$ and $B N A$, it follows that $M N: N B=M C: A B=2: 9$.

Therefore, $S_{M N C}=2 /{ }_{3} \cdot 2 / 11 S_{A D C}=1 / 33 S_{A B C D}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-26.jpg?height=334&width=1054&top_left_y=1705&top_left_x=502)

## Answer

$1: 33$.