Six edges of a tetrahedron fall naturally into 3 pairs of the opposites - the non-intersecting edges. From the point of view of the four vertices, the latter can be split into 2 pairs in three ways. Either way, there are exactly three line segments joining the midpoints of the opposite edges. Three such segments intersect in a point which is the barycenter of the four material points of equal mass placed at the vertices of the tetrahedron. These are called bimedians of the tetrahedron. The applet helps observe the fact that in a regular tetrahedron the bimedians are pairwise orthogonal. It also suggests a simple proof of this fact.

A regular tetrahedron can be embedded into a cube. The bimedians then serve to join the centers of the opposite faces of the cube. By any measure, these lines are pairwise orthogonal.

Obviously, the same holds for a tetrahedron that can be embedded into a parallelopiped. For this to happen, the bimedians need to be orthogonal to the edges they join. This is a generalization of sorts, but it is not very interesting. Posing a problem we in fact stipulate its solution: if the bimedians of a tetrahedron are orthogonal to the edges they join, the three are then pairwise orthogonal. In a regular tetrahedron the required orthogonality is an unannunciated property implied by the embedding into a cube.

References

1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #148