3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. i.e., we have to prove that 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 [SAS congruency] Thus, AB = BC [CPCT] Similarly we can prove, BC = CD CD = AD AD = AB , AB = BC = CD = AD Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.
