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

(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(AB² + BC² + CD²)
AD² + BE² + CF²
3(AB² + BC² + CD²)

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 AD² + BE² + CF² > 3(AB² + BC² + CD²)?
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:
AD² + BE² + CF² ≤ 4(AB² + BC² + CD²)
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(AB² + BC² + CD²) = (AC² + CE² + EA²)
3(AD² + BE² + CF²) = 4(AC² + CE² + EA²).

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:
AB² + BC² + CD² + AC² + CE² + EA² = AD² + BE² + CF²
DE² + EF² + FA² + BD² + DF² + FB² = AD² + BE² + CF² (analogous to the preceding result)
4(AB² + BC² + CD²) - (AD² + BE² + CF²) = 9MN²
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 2n-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:
AB² + BC² + CD² + DE² + EF² + FA² - AC² - BD² - CE² - DF² - EA² - FB² + AD² + BE² + CF² = 9MN²
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 2n 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