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

Points $M, N$, and $K$ belong to the edges $AD$, $AB$, and $BC$ of the tetrahedron $ABCD$, respectively, such that $AM: MD=2: 3$, $BN: AN=1: 2$, and $BK=KC$. Construct the section of the tetrahedron by the plane passing through points $M, N, K$. In what ratio does this plane divide the edge $CD$?
Let the extension of segment $M N$ intersect line $D B$ at point $F$. Then points $F$ and $K$ lie in the plane of face $B C D$, so line $F K$ lies in the plane of face $B C D$. If $P$ is the point of intersection of this line with edge $C D$, then quadrilateral $M N K P$ is the desired section. Let $B D=a, B F=x$. Through vertex $A$, draw a line parallel to $B D$, and extend $M N$ to intersect this line at point $Q$. Triangle $A N Q$ is similar to triangle $B N F$ with a ratio of 2, and triangle $A M Q$ is similar to triangle $D M F$ with a ratio of $\frac{A M}{M D}=\frac{2}{3}$. Therefore,

$$
A Q=2 B F=2 x, A Q=\frac{2}{3} D F=\frac{2}{3}(a+x)
$$

From the equation $2 x=\frac{2}{3}(a+x)$, we find that $B F=\frac{a}{2}$. Through vertex $C$, draw a line parallel to $B D$, and extend $K P$ to intersect this line at point $T$. Triangle $C K T$ is equal to triangle $B K F$, and triangle $C P T$ is similar to triangle $D P F$, so

![](https://cdn.mathpix.com/cropped/2024_05_06_808033c11c6fe49d1738g-39.jpg?height=94&width=568&top_left_y=1142&top_left_x=745)

## Answer

$3: 1$.