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: *AD*² + *BE*² + *CF*² > 3(*AB*² + *BC*² + *CD*²).

Parallelo-hexagon with Obtuse Angles

**Explaining**

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.

**Reference**

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 2*n*-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

Copyright © 2019 KCP Technologies, a McGraw-Hill Education Company. All rights reserved.

Release: 2020Q3, semantic Version: 4.8.0, Build Number: 1070, Build Stamp: 8126303e6615/20200924134412

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