Given : A quadrilateral ABCD, in which diagonals AC and BD bisect each other at right angles.
To Prove : ABCD is a rhombus.
Proof : In ∆AOB and ∆BOC
AO = OC
[Diagonals AC and BD bisect each other]
∠AOB = ∠COB [Each = 90°]
BO = BO [Common]
∴ ∆AOB ≅ ∆BOC [SAS congruence]
AB = BC …(i) [CPCT]
Since, ABCD is a quadrilateral in which
AB = BC [From (i)]
Hence, ABCD is a rhombus.
[∵ if the diagonals of a quadrilateral bisect each other, then it is a parallelogram and opposite sides of a parallelogram are equal] Proved.
