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

Point \( M \) is the midpoint of edge \( AD \) of tetrahedron \( ABCD \). Point \( N \) lies on the extension of edge \( AB \) beyond point \( B \), and point \( K \) lies on the extension of edge \( AC \) beyond point \( C \), such that \( BN = AB \) and \( CK = 2AC \). Construct the section of the tetrahedron by plane \( MNK \). In what ratio does this plane divide the edges \( DB \) and \( DC \)?
Lines $A K, K M, A D$ and $C D$ lie in the same plane. Let $P$ be the point of intersection of the lines $K M$ and $C D$. Draw a line through vertex $D$ parallel to $A K$, and extend $K M$ to intersect this line at point $T$. Let $A C=a$. Then $C K=2 a$. From the congruence of triangles $D M T$ and $A M K$, it follows that $D T=A K=3 a$. From the similarity of triangles $D P T$ and $C P K$, we find that

$$
\frac{\partial P^{\prime}}{P C C}=\frac{N P^{\prime}}{C K}=\frac{3 a}{2 a}=\frac{3}{2}
$$

Similarly, we find that the section plane divides the edge $D B$ in the ratio 2:1, counting from vertex $D$.

## Answer

$2: 1 ; 3: 2$