Question

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

Answer 1

Given,

Diagonals are equal

$AC=BD$ .......$\left(1\right)$

and the diagonals bisect each other at right angles

$OA=OC;OB=OD$ ...... $\left(2\right)$

$\angle AOB=$ $\angle BOC=$ $\angle COD=$ $\angle AOD=$ ${90}^0$ ..........$\left(3\right)$

Proof:

Consider $△AOB$ and $△COB$

$OA=OC$ ....[from $\left(2\right)$]

$\angle AOB=$ $\angle COB$

$OB$ is the common side

Therefore,

$△AOB\cong$ $△COB$

From $SAS$ criteria, $AB=CB$

Similarly, we prove

$△AOB\cong$ $△DOA$, so $AB=AD$

$△BOC\cong$ $△COD$, so $CB=DC$

So, $AB=AD=CB=DC$ ....$\left(4\right)$

So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence $ABCD$ is parallelogram

In $△ABC$ and $△DCB$

$AC=BD$ ...(from $\left(1\right)$)

$AB=DC$ ...(from $(4)$)

$BC$ is the common side

$△ABC\cong$ $△DCB$

So, from SSS criteria, $\angle ABC=$ $\angle DCB$

Now,

$AB\parallel CD,BC$ is the tansversal

$\angle B+\angle C=$ ${180}^0$

$\angle B+\angle B=$ ${180}^0$

$\angle B=$ ${90}^0$

Hence, $ABCD$ is a parallelogram with all sides equal and one angle is ${90}^0$

So, ABCD is a square.

Hence proved.