Assume, for convenience, that $GH = 1$. Then $DH = 1$ and $DG = \sqrt{2}$. Also, $DC = 1$, $BC = \sqrt{3}/3$ because $CG = 1 \text{ and } DB = 2 \sqrt{3}/3.$

Now, $\triangle BDG$ is iscosceles, so when we draw $BM$ to the midpoint of $DG$, we have a right angle at $M$.

So $\triangle BMG$ is a right triangle and now we can nail down the desired cosine.