[ Parallelepipeds (miscellaneous). ] [ Parallelism of lines and planes ]  Through the midpoints $M$ and $N$ of the edges $A D$ and $C C 1$ of the parallelepiped $A B C D A 1 B 1 C 1 D 1$, a plane is drawn parallel to the diagonal $D B 1$. Construct the section of the parallelepiped by this plane. In what ratio does it divide the edge $B B 1$?
Consider a plane passing through the parallel lines $A D$ and $B 1 C 1$. Draw a line through point $M$ in this plane, parallel to $D B 1$. Let $T$ be the point of intersection of this line with the line $C 1 B 1$, and $K$ be the point of intersection of the line $T N$ with the edge $B B 1$. The plane $M N K$ is parallel to the line $D B 1$, since it contains the line $M T$, which is parallel to $D B 1$. Let the line $K N$ intersect the line $B C$ at point $S$, the line $M S$ intersect the edge $C D$ at point $P$, and the line $A B$ at point $L$, and the line $L K$ intersect the edge $A A 1$ at point $Q$. Then the pentagon $M P N K Q$ is the desired section. Let $A D=a$. Then

$$
T B 1=M D=\frac{a}{2}
$$

From the equality of triangles $T N C 1$ and $S N C$, it follows that

$$
C S=T C 1=B 1 C 1+T B 1=\frac{3 a}{2}
$$

and from the similarity of triangles $S B K$ and $T B 1 K-$

$$
B K: K B 1=B S: T B 1=\frac{5 a}{2}: \frac{2}{2}=5: 1
$$

## Answer

$5: 1$.