the joins of tjhe mid point s of the opposite edges of a tetrahedron intersect and bisect each other. proove it.

proof: the computation is easy in the case of a regular tetrahedron inscribed in a cube with edge 1 (each of the four tri-right-angled tetrahedra has a volume of 1/6; thus it remains 1/3 for the regular tetrahedron).

Generalizing is easy: the four pyramids have the same volume (same height, that of the parallelepiped, and bases with same area, half of that of the parallelepiped s base).

The edges of the tetrahedron are six of the twelve diagonals of the parallelepiped s faces.

The six other diagonals of the parallelepiped s faces define a second tetrahedron with same volume. The union of these two tetrahedra is an "anti-parallelepiped" (distorted anticube).