A theorem involving the perpendicular bisectors of a hexagon with opposite sides parallel

Theorem: The intersections of the adjacent perpendicular bisectors of the sides of a hexagon with opposite sides parallel form a parallelo-hexagon, i.e. hexagon with opposite sides parallel and equal.

Alternative formulation: The respective circumcentres G, H, I, J, K and L of triangles ABC, BCD, CDE, DEF and EFA of a hexagon ABCDEF with opposite sides parallel, form a parallelo-hexagon, i.e. hexagon with opposite sides parallel and equal.


1) At first glance, with the number of formed parallelograms, the result may seem to be easy to prove just using the properties of parallelograms. However, the author was unable to do so after several different attempts.1
2) Alternatively, the above theorem seems to lend itself most readily to attack by vector methods or complex algebra, but the algebra gets very complicated rather quickly, so a proof by computer technology such as Mathematica becomes convenient. The real challenge, however, remains: to find a purely geometric proof. Can you find such a proof?

Hint: Click on the Hint button in the sketch for a possible construction which one can use to produce a geometric proof.

Novel Proof
Read my 2020 paper related to this result in the International Journal for Mathematical Education in Science & Technology (IJMEST) which gives a geometric proof using ideas from calculus as well as a Mathematica proof - the 1st 50 downloads are free and complimentary at this link: A novel proof of a hexagon theorem. If no free copies are available any more, please feel free to write to me to request a copy.

1The result is not true for arbitrary perpendiculars to the sides of a general hexagon with opposite sides parallel (as can be easily verified by a rough drawing or accurate construction). This indicates that just using parallel lines, even though they form a number of parallelograms, are not sufficient to prove that the opposite sides of the formed hexagon GHIJKL are equal. The perpendical bisectors all have to intersect at the midpoints of the sides of the original hexagon ABCDEF, i.e. be perpendicular bisectors.

More Proofs
1) I'm aware of a joint paper by German colleagues, Rolfdieter Frank and Heinz Schumann, which gives an alternative proof for the above result, that has been submitted for consideration for publication to IJMEST.
2) Vladimir Dubrovsky from Russia also recently, on 22 November 2021, gave a neat transformation proof for the result in a post at the Romantics of Geometry group on Facebook. Read his proof Dubrovsky proof.

Another result
For another interesting & challenging 'to-prove' result involving a hexagon with opposite sides parallel, go here: Parallel-hexagon concurrency theorem. In this theorem, the lines connecting the midpoints of the opposite sides are concurrent (which in turn is a special case of Toshio Seimiya's Theorem).


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Created by Michael de Villiers, 17 April 2019, updated 8 May 2020; 31 March 2021; 3 April 2021; 29 Nov 2021.