Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
ABCD is parallelogram and AB = BC = CD = AD
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.