**(Relations between the sides and diagonals of parallelo-hexagons, and the general theorem of Douglas - 1981)**

An interesting parallelogram theorem, apparently first noted and proved by Apollonius of Perga (ca. 262 BC - ca. 190 BC), states that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagonals (see Parallelogram Law). It seems natural to consider its possible generalization to a hexagon with opposite sides equal and parallel, i.e. a *parallelo-hexagon*.

**Investigate**

Below is a parallelo-hexagon *ABCDEF*. Drag any of the red vertices and consider the relationships between the following displayed calculations:

4(*AB*² + *BC*² + *CD*²)*AD*² + *BE*² + *CF*²

3(*AB*² + *BC*² + *CD*²)

**Conjecture**

What conjectures can you make? Check if they hold in extreme cases or when *ABCDEF* is concave or crossed.

**Explaining**

Can you EXPLAIN WHY (prove that) your conjectures are true?

Relations between the sides and diagonals of various hexagons

**False Conjecture & Improvement**

From the above investigation, perhaps you also made the plausible conjecture that for a **convex** parallelo-hexagon *AD*² + *BE*² + *CF*² > 3(*AB*² + *BC*² + *CD*²)?

Unfortunately, this conjecture is not generally true. Can you produce a counter-example by using the above dynamic sketch to drag and change the shape of the convex parallelo-hexagon? (Hint: drag the vertices until a pair of opposite angles become very acute.)

However, if the **convex** parallelo-hexagon has **all** its angles obtuse, then the above inequality holds. Click on Parallelo-hexagon with Obtuse Angles to navigate to a separate, accurately constructed dynamic sketch. Use it to explore the inequality.

Can you prove this inequality for a convex parallelo-hexagon with obtuse angles?

**Parallelo-hexagon Inequality**

In the investigation above, however, you should've found that in a hexagon *ABCDEF* with opposite sides equal and parallel, i.e. a *parallelo-hexagon*, the following inequality always holds between its sides and major diagonals:

*AD*² + *BE*² + *CF*² ≤ 4(*AB*² + *BC*² + *CD*²)

Equality holds if, and only if, the main diagonals are parallel to a pair of opposite sides. Click on the '**Link to special parallelo-hexagon**' button above to view an accurately constructed dynamic sketch of this. Drag any of the red vertices to explore.

Note that for a parallelo-hexagon with main diagonals parallel to a pair of opposite sides, the following two additional equalities also hold between its sides and diagonals:

3(*AB*² + *BC*² + *CD*²) = (*AC*² + *CE*² + *EA*²)

3(*AD*² + *BE*² + *CF*²) = 4(*AC*² + *CE*² + *EA*²).

**Parallelo-hexagon Equalities**

In a hexagon *ABCDEF* with opposite sides equal and parallel, i.e. a general *parallelo-hexagon*, with *M* and *N* the respective centroids of triangles *ACE* and *BDF*, the following three relationships hold between its sides, diagonals and the distance *MN*:

*AB*² + *BC*² + *CD*² + *AC*² + *CE*² + *EA*² = *AD*² + *BE*² + *CF*²

*DE*² + *EF*² + *FA*² + *BD*² + *DF*² + *FB*² = *AD*² + *BE*² + *CF*² (analogous to the preceding result)

4(*AB*² + *BC*² + *CD*²) - (*AD*² + *BE*² + *CF*²) = 9*MN*²

Click on the '**Link to special Douglas theorem**' button above to view an accurately constructed dynamic sketch displaying these relationships. Drag any of the red vertices to explore. Also check concave and crossed cases.

**The theorem of Douglas**

Remarkably, the last result above can be viewed as a special case of **Douglas' Theorem** (1981) for 2*n*-gons in general. For a general hexagon *ABCDEF* with *M* and *N* the respective centroids of triangles *ACE* and *BDF*, according to the theorem of Douglas, the following relationship holds between its sides, diagonals and the distance *MN*:

*AB*² + *BC*² + *CD*² + *DE*² + *EF*² + *FA*² - *AC*² - *BD*² - *CE*² - *DF*² - *EA*² - *FB*² + *AD*² + *BE*² + *CF*² = 9*MN*²

Click on the '**Link to general Douglas theorem**' button above to view an accurately constructed dynamic sketch displaying this relationship for a general hexagon. Drag any of the red vertices to explore. Also check concave and crossed cases.

**References**

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

De Villiers, M. (2013). Why still factorize algebraic expressions by hand? *Learning & Teaching Mathematics*, no. 14, pp. 44-46.

Douglas, A.J. (1981). A generalization of Apollonius' theorem. *Mathematical Gazette*, 65(431), pp. 19-22.

Note that that the 2*n* points in Douglas' generalization do not have to be co-planar, but can be in 3D space. Also note that what he defines in his paper as 'orthocentres' are normally called 'centroids'; i.e. since they are obtained by the arithmetic average of the coordinates of the points.

**Related Links**

Parallelo-hexagon with Obtuse Angles

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

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Created, Michael de Villiers, 9 April 2013; updated to *WebSketchpad*, 19 Sept 2023; updated 21 Sept 2023.