Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and âˆ AOB = âˆ BOC = âˆ COD = âˆ AOD = 90Âº.

Now, to prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90Âº.

**Proof:**

In Î”AOB and Î”COD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

âˆ AOB = âˆ COD (Vertically opposite angles)

âˆ´ Î”AOB â‰… Î”COD (SAS congruence rule)

âˆ´ AB = CD (By CPCT) ... (1)

And, âˆ OAB = âˆ OCD (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.

âˆ´ AB || CD ... (2)

From equations (1) and (2), we obtain

ABCD is a parallelogram.

In Î”AOD and Î”COD,

AO = CO (Diagonals bisect each other)

âˆ AOD = âˆ COD (Given that each is 90Âº)

OD = OD (Common)

âˆ´ Î”AOD â‰… Î”COD (SAS congruence rule)

âˆ´ AD = DC ... (3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)

âˆ´ AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In Î”ADC and Î”BCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

âˆ´ Î”ADC â‰… Î”BCD (SSS Congruence rule)

âˆ´ âˆ ADC = âˆ BCD (By CPCT)

However, âˆ ADC + âˆ BCD = 180Â° (Co-interior angles)

â‡’ âˆ ADC + âˆ ADC = 180Â°

â‡’ 2âˆ ADC = 180Â°

â‡’ âˆ ADC = 90Â°

One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90Âº. Therefore, ABCD is a square.