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In the given figure, prove that:
(i) ∆ AOD ≅ ∆ BOC
(ii) AD = BC
(iii) ∠ADB = ∠ACB
(iv) ∆ADB ≅ ∆BCA

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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.

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