Draw the line AC, it will cross DB at the middle because of the symetry of the parallelogram. Consider the two triangles BEA and ECA, their baces is the same (along the line BC) and so is their heights, hence also their area.

Now consider the two triangles BEG and ECG, by the same reason they have equal area as well. That means the two triangles GCA and GBA also has the same area, cause GCA + ECG = ECA, and GBA + BEG = BEA.

Now, also AC is split on the middle by BD, because of the symetry. If the midpoint (where they cross) is M, that means also triangles AMG and CMG has the same area, hence half the area of ACG. And since ACG and ABG has the same area, AMG has half the area of ABG. If we say their bases is along the line the line BM, GMA has the same height as BGA, hence the half base, hence GM is half the distance of BG and one third of BM.

If you do the same thing on the other half of the parallelogram, you'll se HM is one third of DM. And by adding two and two together you'll se DH equals HG equals GB.

Hope it's readable

/Kristofer