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Show thal 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.