Given a quadrilateral pyramid SABCD, the base of which is a trapezoid ABCD. The ratio of the bases AD and BC of this trapezoid is 2. Construct the section of the pyramid by a plane passing through point $D$ and the midpoints of edges SA and SB. In what ratio does this plane divide the edge SC?

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Let M and N be the midpoints of edges SA and SB, respectively. According to the theorem of intersecting planes through two parallel lines, the planes of faces $ASD$ and $BSC$ intersect along a line parallel to $AD$ and $BC$. Let $K$ be the point of intersection of this line with line $DM$, and $L$ be the point of intersection of lines $KN$ and $SC$. Then the quadrilateral $DMNL$ is the desired section. Let $BC = a$. Then $AD = 2a$. From the congruence of triangles KMS and DMA, it follows that $KS = AD = 2a$. Extend $KL$ to intersect line $BC$ at point $P$. From the congruence of triangles $BNP$ and $SNK$, we find that $BP = KS = 2a$. Therefore,

$$
CP = BP - BC = 2a - a = a \text{. }
$$

Finally, from the similarity of triangles KLS and PLC, we find that

$$
\frac{KL}{LC} = \frac{KS}{CP} = \frac{2a}{a} = 2
$$

## Answer

$2: 1$