Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.
In ΔAOD and ΔCOD,
OA = OC (Diagonals bisect each other)
∠AOD = ∠COD (Given)
OD = OD (Common)
∴ ΔAOD ≅ ΔCOD (By SAS congruence rule)
∴ AD = CD (1)
Similarly, it can be proved that
AD = AB and CD = BC (2)
From equations (1) and (2),
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram.
Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
