Let us prove that, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square.

Consider the parallelogram ABCD with diagonals AC and BD as shown and they intersect at right angles at O

A parallelogram is a square if its adjacent angles are 90° and adjacent sides are equal

So we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a square

Consider ΔBAD and ΔCDA

AC = BD … given

AB = DC … opposite sides of a parallelogram

AD is the common side

Therefore, ΔBAD ≅ ΔCDA … SSS test for congruency

⇒ ∠BAD = ∠CDA …corresponding angles of congruent triangles …(i)

As it is given that ABCD is parallelogram

⇒ ∠BAD + ∠CDA = 180° … sum of adjacent angles of a parallelogram is 180°

Using equation (i)

⇒ ∠CDA + ∠CDA = 180°

⇒ 2 × ∠CDA = 180°

⇒ ∠CDA = 90°

⇒ ∠BAD = 90°

Thus, adjacent angles are 90° … (iii)

Consider ΔAOB and ΔAOD

∠AOB = ∠AOD … both 90° because given that diagonals intersect at right angles

OD = OB … diagonals of a parallelogram bisect each other

AO is the common side

Therefore, ΔAOB ≅ ΔAOD … SAS test for congruency

⇒ AB = AD … corresponding sides of congruent triangles

Thus, adjacent sides are equal … (iv)

From statements (iii) and (iv) we can conclude that parallelogram ABCD is a square

Therefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square.