Parallelo-hexagon with Obtuse Angles

Below is shown a dynamic sketch of a hexagon with opposite sides equal and parallel, i.e. a parallelo-hexagon. More-over, it has been constructed in such a way that all its angles are obtuse. Explore the sketch by dragging any of the red vertices or the points H and I to change the obtuse angles at vertices A and F.
Theorem: If ABCDEF is a convex parallelo-hexagon with all its angles obtuse, then the following inequality holds: AD² + BE² + CF² > 3(AB² + BC² + CD²).

Parallelo-hexagon with Obtuse Angles

Can you EXPLAIN WHY (prove that) this theorem is true?

Explore More
To explore more about the relations between the sides and diagonals of parallelo-hexagons, or general hexagons, click on Relations between the sides and diagonals of parallelo-hexagons, and the general theorem of Douglas to navigate to a separate, accurately constructed dynamic sketch.

De Villiers, M. (2012). Relations between the sides and diagonals of a set of hexagons. Mathematical Gazette, July, 96(536), pp. 309-315.

Related Links
A generalization of a Parallelogram Theorem to Parallelo-hexagons, Hexagons and 2n-gons in general
Easy Hexagon Explorations
The 3D parallelo-hexagon
Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem
Some Parallelo-hexagon Area Ratios
Area Parallelogram Partition Theorem: Another Example of the Discovery Function of Proof

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Created, Michael de Villiers, 21 September 2023 using WebSketchpad.