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Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

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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.

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