The base of the pyramid $S A B C D$ is a parallelogram $A B C D$. A plane is drawn through the side $A B$ and the midpoint $M$ of the lateral edge $S C$. 1) Construct the section of the pyramid by this plane. 2) In what ratio does this plane divide the volume of the pyramid?
1) The secant plane and the plane of the face $DSC$ pass through the parallel lines $AB$ and $CD$ respectively and have a common point $M$. Therefore, they intersect along a line passing through the point $M$ and parallel to $AB$ and $CD$. If $N$ is the point of intersection of this line with the lateral edge $SD$, then $MN$ is the midline of the triangle $DSC$.

Thus, the desired section is a trapezoid $ABMN$, in which the base $MN$ is half the length of the base $AB$ ( $MN = \frac{1}{2} AB = \frac{1}{2} CD$ ).

2) Let $V$ be the volume of the given pyramid $SABCD$. Then the volume of each of the two triangular pyramids $SABD$ and $SCBD$ is $\frac{1}{2} V$ (these pyramids have equal bases $ABD$ and $CDB$ and a common height drawn from their common vertex $S$). The triangular pyramids $NABD$ and $SABD$ have a common base $ABD$, and the height of the pyramid $NABD$, drawn from the vertex $N$, is half the height of the pyramid $SABD$, drawn from the vertex $S$, since $N$ is the midpoint of the edge $SD$. Therefore,

$$
V_{NABD} = \frac{1}{2} V_{SABD} = \frac{1}{4} V
$$

Thus, the volume of the triangular pyramid $SANB$ is also $\frac{1}{4} V$. The area of the base $SMB$ of the triangular pyramid $NSMB$ is half the area of the base $SBC$ of the triangular pyramid $DSBC$ (since $BM$ is the median of the triangle $SBC$), and the height of the pyramid $NSMB$, drawn from the vertex $N$, is half the height of the pyramid $DSBC$, drawn from the vertex $D$. Therefore, the volume of the pyramid $NSMB$ is one-fourth the volume of the pyramid $DSBC$, i.e.,

$$
V_{NSMB} = \frac{1}{4} V_{DSBC} = \frac{1}{8} V.
$$

The volume of the quadrilateral pyramid $SABMN$, cut off by the secant plane from the given pyramid $SABCD$, is the sum of the volumes of the triangular pyramids $SANB$ and $NSMB$, i.e.,

$$
V_{SABMN} = \frac{1}{4} V + \frac{1}{8} V = \frac{3}{8} V.
$$

Therefore, the secant plane divides the volume of the pyramid $SABCD$ in the ratio $\frac{3}{5}$.

## Answer

$3: 5$