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

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Given :

Diagonals of the quadrilateral bisect each other at right angles.

To do :

We have to show that it is a rhombus.

Solution:

Let $ABCD$ be a quadrilateral in which diagonals bisect each other at right angles.

So,

$OA=OC, OB=OD$

$\angle AOB = \angle BOC =\angle COD =\angle AOD = 90^o$

To prove that it is a rhombus, we need to prove that $ABCD$ is a parallelogram and $AB = BC = CD = AD$

In $\triangle AOB$ and $\triangle BOC$,

$OA=OC$        (Given)

$OB=OB$        (Common)

$\angle AOB= \angle BOC$  ($90^o$)

Therefore, by SAS congruency, we get,

$\triangle AOB \cong \triangle BOC$

So, $AB=BC$          (CPCT)

Similarly,

$\triangle AOB \cong \triangle AOD$

So, $AB=AD$

$\triangle COD \cong \triangle BOC$

So, $CD=BC$

Therefore,

$AB=BC=CD=AD$

We can say that,

$AB=CD$ and $BC=AD$

As the opposite sides are equal, $ABCD$ is a parallelogram.

We know that a parallelogram whose diagonals intersect at right angles is a rhombus.

Therefore, $ABCD$ is a parallelogram with all sides equal and diagonals bisect each other at right angles.

So, $ABCD$ is a rhombus.

Hence proved.

Updated on 10-Oct-2022 13:40:58