**Theorem**

The intersections of the adjacent perpendicular bisectors of the sides of a *parallel-hexagon* (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.

**Challenge**

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 novel, geometric proof.

**Novel Proof**

Read my 2020 online 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 at: A novel proof of a hexagon theorem.

**Note**:
^{1}The 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 perpendiculars all appear to have to intersect at the midpoints of the sides of the original hexagon *ABCDEF*, i.e. be perpendicular bisectors.

**More Proofs**

More recently, the following more elementary & explanatory proofs have been given:

Dubrovsky, V. (2021). Dubrovsky proof, given on 22 November 2021, in a post at the *Romantics of Geometry* group on Facebook.

Frank, R. & Schumann, H. Elementary proof of a theorem about hexagons with parallel opposite sides. *International Journal of Mathematical Education in Science and Technology*, Published online: 19 Oct 2023.

**Another Result**

For another interesting & challenging 'to-prove' result involving a *parallel-hexagon* (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).

**Related Links**

Conway's Circle Theorem as special case of Side Divider Theorem

Area ratios of some polygons inscribed in quadrilaterals and triangles

Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem

Some Parallelo-hexagon Area Ratios

A side trisection triangle concurrency

A generalization of a Parallelogram Theorem to Parallelo-hexagons, Hexagons and 2*n*-gons in general

A 1999 British Mathematics Olympiad Problem and its dual

Easy Hexagon Explorations

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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; 25 October 2023.