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

Given a tetrahedron $A B C D$. Points $M, N$, and $K$ lie on the edges $A D, B C$, and $D C$ respectively, such that $A M: M D=1: 3$, $B N: N C=1: 1$, and $C K: K D = 1: 2$. Construct the section of the tetrahedron by the plane $M N K$. In what ratio does this plane divide the edge $A B$?
Let $F$ be the intersection point of lines $KM$ and $CA$ (Fig.1), and $P$ be the intersection point of lines $FN$ and $AB$. Then the quadrilateral $MKNP$ is the desired section. Draw a line through vertex $D$ parallel to $AC$, and extend $MK$ to intersect this line at point $T$ (Fig.2). Let $AC = a$, $AF = x$. From the similarity of triangles $DKT$ and $CKF$, we find that

$$
DT = FC \cdot \frac{DK}{KC} = 2(a + x)
$$

and from the similarity of triangles $DMT$ and $AMF$,

$$
DT = AF \cdot \frac{DM}{MA} = 3x.
$$

From the equation $2(a + x) = 3x$, we find that $AF = x = 2a$. Draw a line through vertex $B$ parallel to $AC$, and extend $FN$ to intersect this line at point $E$ (Fig.3). From the equality of triangles $BNE$ and $CNF$, it follows that $BE = CF = 3a$, and from the similarity of triangles $BRE$ and $APF$,

$$
\frac{AP'}{P'B} = \frac{AF'}{BE'} = \frac{2a}{3a} = \frac{2}{3}
$$

## Answer

$2: 3$