Answer 1

Proof:

In Δ AOD and Δ BOC 

OA = OB ........(given)

∠AOD = ∠BOC  .............(vertically opposite angles)

OD = OC ............(given)

(i) ∴ Δ AOD ≅ Δ BOC ................(S.A.S. Axiom)

Hence (ii) AD = BC ..........(c.p.c.t.)

and (iii) ∠ADB = ∠ACB .....(c.p.c.t.)

(iv) Δ ADB ≅ Δ BCA

Δ ADB = Δ BCA ..............(Given)

AB = AB .................(Common)

∴ Δ AOB ≅ Δ BCA

Hence proved.