Exercise:
Prove that a tetrahedral angle can be intersected by a plane in such a way that the cross section is a parallelogram.
Attempt:
Draw an arbitrary parallelogram $ABCD$ on an arbitrary plane $P$.
Pick every 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 intersecting $P$.
Pick every point $T$ on $P$.
$TEFGH$ is a tetrahedral angle with vertex $T$ with a cross section of a parallelogram $EFGH$.
Every point $S$ and $T$ covers all points.
So, every point can be a vertex of a tetrahedral angle that has a cross section of a parallelogram.
Request:
I realized my attempt doesn't exactly prove what is asked, so it isn't valid. However, I've share it anyways to show that I have tried to do something about this exercise, and maybe I had the right idea.
I think that the most reasonable proof would be proof-by-construction, however I'm not sure how to go about such proof. Hints and solutions are welcome.