given parallelogram ABCD, EF bisects AB, EF bisects CD at F Prove that Eg is congruent to FG

Question

Must go to equality to get answer

Answer 1

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.