**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.**