Another Example of the Discovery Function of Proof

In my 1990 paper on proof the 'discovery' function of proof was illustrated with an example of where further reflection on an 'explanatory' proof leads to a new more general result. This illustrates that understanding why a result is true is an added benefit of proof. Below is another example of an elementary result that can easily be used in the classroom to illustrate this discovery function of proof.

The following lovely little result is given in a paper by Philippe Richard (2003) and was also used as a problem in one of the Intermediate Olympiads in the UK (Jobbings, 2013): Given a parallelogram ABCD and arbitrary points E and F, then as shown below, Area BGE + Area IFD + Area ECFH = Area AGHI.
Select and drag any of A, B, C, E or F to dynamically change the figure and investigate the result.

Challenge:
1) Can you explain why (prove that) the above result is true?
2) Click on the Show Hint button if necessary.
3) If still stuck, first try this simpler, auxiliary problem "Simpler Partition of a Parallelogram", to prove the equality of areas EGH + DIC = AGD + HIF.
4) Carefully reflect on your proof(s), and consider how this same proof can also apply to a certain type of pentagon, hexagon, etc. Make generalizations and check your generalized conjectures by clicking on the Link buttons on the right to go to pentagons, hexagons, etc. with a similar area partition property.
5) Can you explain (prove) these generalizations as well?
6) Is it necessary for the area partition result to hold for a 2n-gon with n > 2 that the 2n-gon must have all pairs of opposite sides equal? Explain why, or why not?