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

Given :

Diagonals of the quadrilateral are equal and bisect each other at right angles.

To do :

We have to show that it is square.

Solution :

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

So, $AC=BD$

$OA=OC, OB=OD$

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

To prove that it is a square, we need to prove the quadrilateral is a parallelogram and one of the angles is $90^o$.

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

$OA=OC$ (Given)

$OB=OB$ (Common)

$\angle AOB= \angle BOC$ ($90^o$)

Therefore,

$\triangle AOB \cong \triangle BOC$

So, $AB=BC$

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.

In $\triangle ABC$ and $\triangle BCD$,

$AB=CD$ (Opposite sides of the parallelogram)

$BC=BC$ (Common)

$AC=BD$ (Diagonals are equal)

Therefore,

$\triangle ABC \cong \triangle BCD$

So, $\angle B= \angle C$

We know that adjacent angles of a parallelogram are supplementary.

$\angle B + \angle C = 180^o$

$\angle B= \angle B = 180^o$

$2 \angle B= 180^o$

$\angle B = \frac{180^o}{2}$

$\angle B =90^o$

Therefore, $ABCD$ is a parallelogram with all sides equal and one angle is $90^o$.

So, $ABCD$ is a square.

Hence proved.

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