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.
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.