**(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*. Below are some interesting properties related to such an investigation.

1) In a hexagon *ABCDEF* with opposite sides equal and parallel, i.e. a *parallelo-hexagon*, and the diagonals are parallel to a pair of opposite sides, the following three relationships hold between its sides and diagonals:

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

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)

Consider the accurately constructed dynamic sketch below. Drag any of the red vertices to explore.

Fig. 1 Parallelo-hexagon with Diagonals parallel to Opposite sides - Three equalities

2) 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*:

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

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

4(*AB*² + *BC*² + *CD*²) - (*AD*² + *BE*² + *CF*²) = 9*MN*² (This result is a special case of the theorem of Douglas - see no. 3 below)

Consider the accurately constructed dynamic sketch below. Drag any of the red vertices to explore. Also check concave and crossed cases.

Fig. 2 Parallelo-hexagon with Centroids - One inequality and two equalities

3) Remarkably, all the preceding results are merely special cases of **Douglas' Theorem** (1981) for 2*n*-gons. For a general hexagon *ABCDEF* with *M* and *N* the respective centroids of triangles *ACE* and *BDF*, 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

Consider the accurately constructed dynamic sketch below. Drag any of the red vertices to explore. Also check concave and crossed cases.

Fig 3 Douglas' theorem (1981) for sides and diagonals of 2*n*-gons

On online investigation for students is available in the section for *Explorations for Students*. Click Here to go there.

Read my paper in the July 2012 issue of the *Mathematical Gazette* at: Relations between the sides and diagonals of a set of hexagons.

Read Douglas' paper in the March 1981 issue of the *Mathematical Gazette* at: A generalization of Apollonius' theorem. Perhaps even more remarkable, is 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' is normally called 'centroids'; i.e. since they are obtained by the arithmetic average of the coordinates of the points.

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Created by Michael de Villiers, 8 April 2013.