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)
