Matthew M. Conroy, University of Washington
One of Van Aubel's theorem's says that if you append squares to the sides of a quadrilateral, then the line segments connecting the centers of those squares are perpendicular to each other and of equal length.
An extension of this theorem to octagons says that if we append squares to the sides of an octagon, and connect the opposite centers of the squares with line segments, and then connect the midpoints of those line segements, the two resulting segments are perpendicular and of equal length. To be clearer, if the centers (walking around the octagon) are p0,p1,...,p7, we find the midpoints of p0p4, p1p5, p2p6, and p3p7; call these midpoints m,n,o,p. Then the line segments mo and np are perpendicular and of equal length.
I do not know a proof of this result, or its history and only came upon it via Dao Thanh Oai's paper which mentions it without proof.
If we instead create a new octagon from the original by connecting the midpoints of its sides, and then apply the process described above, the resulting four points form a square. This theorem is called Thebault's theorem in Dao Thanh Oai's paper. It is discussed a bit here.
Just for fun, we can apply the midpointing process again. Of course, the result yields squares. I wonder what properties these squares might have?