If the diagonals of a rectangle intersect each other at right angle, the rectangle is a square.

Given: ABCD is a rectangle such that diagonals BD and AC intersect at O.

To prove: The given rectangle is a square. (AB = BC = CD = AD)

Proof: ∠AOB = 90°, ∠AOD = 90°

Since, opposite sides of rectangle are equal.

∴ AB = DC and AD = BC

In Δ AOB and Δ AOD,

AO = AO (Common Side)

∠AOB = ∠AOD (Both are 90°)

OB = OD (Diagonals bisect each other)

Using SAS congruency criterion,

Δ AOB ≅ Δ AOD

By corresponding parts of congruent triangles,

AD = AB

As we know, AB = DC and AD = BC,

⇒ AB = BC = CD = DA

Since, all sides are equal and diagonals intersect at right angles.

Hence, the rectangle is a square.