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