Given,

ABCD is a rhombus,

EABF is a straight line

EA = AB = BF

OA = OC

OB = OD (diagonals of rhombus are perpendicular bisector of each other)

∠AOD = ∠COD = 90^∘

∠AOB = ∠COB = 90^∘

In ΔBDE,

A and O are mid points of BE and BD

OA ⎸⎸DE

OC ⎸⎸DG

In ΔCFA

B and O are the mid points of AF and AC

OB ⎸⎸CF and

OD ⎸⎸GC

Thus in quadrilateral DOCG

OC ⎸⎸DG and

OD ⎸⎸GC

DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 90 (proved)