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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given, OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show, ABCD is parallelogram and AB = BC = CD = AD Proof, 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.