If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

Let ABCD be a cyclic quadrilateral with AD = BC.

We need to prove that AC = DB.

In triangles AOD and BOC:

AD = DB (given)

∠OAD = ∠OBC and ∠ODA = ∠OCB (angle subtended by the same segment are equal)

∴ ΔAOD ≅ ΔBOC (By ASA congruence rule)

⇒ ΔAOD + ΔDOC ≅ ΔBOC + ΔDOC (adding a similar quantity on both sides)

⇒ ΔADC ≅ ΔBCD

∴ AC = BD (by CPCT).

Hence, proved.