Show that the diagonals of a square are equal and bisect each other at right angles.

Let ABCD be the square.

In Δ ABC and Δ BAD,

⇒ AB = AB (Common side)

⇒ BC = AD (Each side of a square is equal in length)

⇒ ∠ABC = ∠BAD = 90° (Each interior angle in a square is a right angle)

∴ Δ ABC ≅ Δ BAD (By S.A.S. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

∴ AC = BD (By C.P.C.T.)

In Δ OAD and Δ OCB,

⇒ AD = CB (Each side of a square is equal in length)

⇒ ∠OAD = ∠OCB (Alternate angles are equal)

⇒ ∠ODA = ∠OBC (Alternate angles are equal)

∴ Δ OAD ≅ Δ OCB (By A.S.A. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ OA = OC (By C.P.C.T.) ……(1)

⇒ OB = OD (By C.P.C.T.) …..(2)

In Δ OBA and Δ ODA,

⇒ OB = OD ….[From equation (2)]

⇒ OA = OA (Common side)

⇒ BA = DA (Each side of a square is equal in length)

∴ Δ OBA ≅ Δ ODA (By S.S.S. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ ∠AOB = ∠AOD = x (let) (By C.P.C.T.)

⇒ ∠AOB + ∠AOD = 180° (Linear pair)

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

⇒ ∠AOB = 90° and ∠AOD = 90°.

Thus, AC and BD bisect each other at right angles.

Hence, proved that the diagonals of a square are equal and bisects each other at right angles.