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