$\left.\begin{array}{l}{[\quad \text { Properties of sections }} \\ \text { [Theorems of Ceva and Menelaus }]\end{array}\right]$

On the edges $A B, B C$, and $A D$ of the tetrahedron $A B C D$, points $K, N$, and $M$ are taken respectively, such that $A K: K B = B N: N C = 2: 1$, and $A M: M D = 3: 1$. Construct the section of the tetrahedron by the plane passing through points $K, M$, and $N$. In what ratio does this plane divide the edge $C D$?
Let $P$ be the point of intersection of the lines $K N$ and $A C$ (Fig.1). Points $P$ and $M$ belong to the planes $M N K$ and $A C D$, so $M P$ is the line of intersection of these planes. Let $F$ be the point of intersection of the lines $M P$ and $C D$. Then the quadrilateral $M K N F$ is the desired section. Consider the plane of triangle $A B C$ (Fig.2). Through vertex $B$, draw a line parallel to $A C$. Let $T$ be the point of intersection of the drawn line with the line $N K$. From the similarity of triangles $T N B$ and $P N C$, it follows that

$$
B T = C P \cdot \frac{B N}{N C} = 2 C P,
$$

and from the similarity of triangles $T K B$ and $P K A$,

$$
B T = A P \cdot \frac{B K}{A K} = \frac{1}{2} A P
$$

thus,

$$
C P = \frac{1}{4} A P, \quad C P = \frac{1}{3} A C.
$$

Now consider the plane of triangle $A D C$ (Fig.3). Reasoning similarly, we get that

$$
\frac{F F}{F C} = \frac{4}{3}.
$$

## Answer

$4: 3$.

## Problem