There seems to be some info missing, so I make some assumptions: EF bisects AB at E and CD at F; the diagonals cross at G (G is not defined in the question). The vertices of the parallogram in clockwise order are ABCD.
BAC=ACD, alternate angles between parallel lines.
AE=EB=DF=FC, because EF is a bisector and the opposite sides of a parallelogram are equal in length.
In triangles AEG and CFG:
angles EAG and GCF are equal because EAG=BAC and GCF=ACD;
angles AEG and GFC are equal, alternate angles between parallel lines;
AE=FC.
The triangles AEG and CFG are congruent (AAS, two angles and a side), therefore EG=FG.