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).
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