Bimedians in a Regular tetrahedron
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.
For a regular tetrahedron, because of the symmetry, the angles between the bimedians may be expected to be "regularly distributed" - they should not change under some simple transofrmations of the tetrahedron. In fact, in a regular tetrahedron all the angles between the three bimedians are right - 90°.
While the claim is quite natural, the proof may not be just one step. However, embedding a regular polyhedron into a cube solves the problem immediately: the bimedians in the tetrahedron join the opposite sides of the cube:
If this is not a proof without words, I do not know what is, but, of course, an interactive illustration beats static graphics hands down.
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