I drew a rough sketch of the figure on paper. (I do't know how to draw/graph in computers. Knowing how to do that will surely help a lot in solving geometry problems, for those drawings are exact.)

So triangle FCG is an equilateral triangle because EF = EC = EG.

In triangles FAD and FCG:

FA is parallel to FG,

FD is parallel to FC,

AD is parallel to CG.

Therefore, triangles FAD and FCG are similar....and so are equilateral both.

So, AD = DF.

If a line is drawn through point F, parallel to AD or BC, this line will interesect AB at, say, point H.

Then, in parallelogram ADFH, AD = DF = FH = HA.

ADFH then is a rhombus.

AF is a segment of the .

AF is a diagonal of rhombus ADFH.

AF bisects angle DAB then.

Therefore, bisects angle DAB also ----proven.

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If by not through the characteristics of a rhombus, AF can be proven as bisecting angle DAB by considering both triangles DAF and HAF are equilateral (and equiangular) triangles. That angle DAF = angle HAF = 60 degrees. Etc.