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

Construct the section of the triangular pyramid $A B C D$ by a plane passing through the midpoints $M$ and $N$ of edges $A C$ and $B D$ and point $K$ on edge $C D$ such that $C K: K D=1: 2$. In what ratio does this plane divide the edge $A B$?
Let the lines $KM$ and $AD$ intersect at point $R$ (Fig.1). Then point $R$ lies on the line of intersection of the cutting plane with the plane of face $ADB$ (since $R$ is a common point of lines lying in these planes). Let the line $RN$ intersect the edge $AB$ at point $L$. Then $L$ is the point of intersection of the cutting plane with the edge $AB$. Therefore, the desired section is the quadrilateral $KMLN$. Draw a line through vertex $C$, parallel to $DR$, until it intersects the line $MK$ at point $P$ (Fig.2). From the similarity of triangles $DKR$ and $CKP$ (by two angles), it follows that

$$
DR = CP \cdot \frac{DK}{KC} = 2CP,
$$

and from the equality of triangles $AMR$ and $CMP$ it follows that $AR = CP$. Therefore, $AR = \frac{1}{2} DR$, i.e., $A$ is the midpoint of $DR$. Draw a line through vertex $B$, parallel to $DR$, until it intersects the line $RN$ at point $Q$ (Fig.3). From the equality of triangles $BNQ$ and $DNR$ it follows that $BQ = DR$, and from the similarity of triangles $ALR$ and $BLQ$ we find that

$$
\frac{AL}{LB} = \frac{AR}{BQ} = \frac{1}{2} \cdot \frac{DR}{DR} = \frac{1}{2}
$$

## Answer

$1: 2$.