Question
In the given figure, prove that:
(i) $\triangle AOD \cong \triangle BOC$
(ii) $A D=B C$
(iii) $\angle ADB =\angle ACB$
(iv) $\triangle ADB \cong \triangle BCA$

Answer

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.

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