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Question
In the given figure, prove that:
- ∆AOD ≅ ∆ BOC
- AD = BC
- ∠ADB = ∠ACB
- ∆ADB ≅ ∆ BCA
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Solution
Proof:
In ΔAOD and ΔBOC
OA = OB ...(given)
∠AOD = ∠BOC ...(vertically opposite angles)
OD = OC ...(given)
(i) ∴ ΔAOD ≅ ΔBOC ...(S.A.S. Axiom)
Hence (ii) AD = BC ...(c.p.c.t.)
and (iii) ∠ADB = ∠ACB ...(c.p.c.t.)
(iv) ΔADB ≅ ΔBCA
ΔADB = ΔBCA ...(Given)
AB = AB ...(Common)
∴ ΔAOB ≅ ΔBCA
Hence, proved.
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