Proof:
In △ AOD and △ BOC
OA = OB (given)
∠AOD = ∠BOC (vertically opposite angles)
OD = OC (given)
(i) ∴ △ AOD ≅ △BOC (S.A.S. Axiom)
Hence (ii) AD = BC (c.p.c.t.)
and (iii) ∠ADB = ∠ACB (c.p.c.t.)
(iv) △ADB ≅ △BCA
△ADB = △BCA (Given)
AB = AB (Common)
△AOB ≅ △BCA
Hence proved.