Question

Solid Geometry. Problems on Proof.
Prove that every convex tetrahedral angle can be cut by a plane so that a parallelogram results in the section.

Answer 1

Above is a parallelogram $ABCD$ on a plane $P$. We have to pick a point $S$ not on $P$

$SABCD$ is a tetrahedral angle with vertex $S$ with a cross section of a parallelogram $ABCD$ draw another arbitrary parallelogram $EFGH$ not interesting $P$

Pick every point $T$ on $P$

$TEFGH$ is a tetrahedral angle with vertex $T$ with a cross section of a parallelogram $EFGH$ every point sand $T$ cover all points.

So, Every point can be a vertex of a tetrahedral angle that has a cross section of parallelogram.