The base of the pyramid SABCD is a parallelogram ABCD. What figure is obtained by the section of this pyramid by the plane $ABM$, where $M$ is a point on the edge SC?
Let K be the point of intersection of the secant plane with the edge SD. The planes ABM and CSD are drawn through the parallel lines $AB$ and $CD$ and intersect along the line KM, so the line KM is parallel to each of these lines, in particular, KM $\|$ AB. Thus, in the section, a quadrilateral $ABMK$ is obtained, in which $KM \| AB$, and $KM < CD = AB$. Therefore, ABMK is a trapezoid.

## Answer

Trapezoid.