Vertex being the center of a circle proving Prove that G is a center of the circle circumscribed on BED.

So we know that we should prove that G lies on the perpendicular bisectors of every side of the BED triangle. For |ED| we can note that $\triangle EDG$ is equilateral so G, as the opposite vertex of |ED|, has to be on the perpendicular bisector.
And... I'm stuck. How can I show it for the other two? There is a ton of angles here but I don't know how to put them to use. Could you tell me what I should take into account while dealing with geometry problems like this? I don't know what to look at and the amount of information on the picture kind of makes me dizzy and everything looks useful and useless at the same time.
 A: Note that $EC=ED$ (isosceles triangle) and therefore $EC=EG$. So it is enough to show that $BG=EC$, for that will show that  $G$ is equidistant from all three vertices of $\triangle EDB$.
Let $DC$ meet $EG$ at $X$, and let $DB$ meet $EG$ at $Y$. Join $CG$. Use angle-chasing to show that parallelograms $EBCY$ and $BCGX$ are congruent. Then $BG$ and $EC$ are corresponding diagonals of congruent parallelograms, so they are equal.   
Remark: One can make the proof more attractive by extending $DG$ so that it meets $AC$ extended. Join $CG$ and $BG$.  Now we have a very nicely symmetric picture, and a solution writes itself down.
A: I think I got it!
Let the brown line cross EG at X, red cross EG at Y. Then EDX is congruent to DGY (the same angles). Also, $\angle GED=60^\circ$ and, as $\angle DAC = 60^\circ$ as well, EG is parallel to AC. Let Z be a point on CD on the same level as F forming a line FZ parallel to EG and AC. Then BZ=CF (BDC is isosceles) and FE=ZG (altitudes of the congruent triangles). Hence, BZ+ZG=BG=EG=DG q.e.d.
