At least one vertex of a tetrahedron projects to the interior of the opposite triangle How i can give a fast proof of the following fact:
Given four points on $\mathbb{R}^3$ not contained in a plane we can choose one such that its projection to the plane passing through the others is in the triangle generated by the three others points.
Thanks in advance.

Here is a graphical supplement (that I cannot place into a comment) to the excellent answer by @achille hui .
I have taken the case $\eta=0.4.$ with normals in red.
The (complicated) name of this polyhedron is "tetragonal disphenoid" (https://en.wikipedia.org/wiki/Disphenoid).

