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Question
In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 
Solution
In triangles AOE and COD,
∠A = ∠C ...(given)
∠AOE = ∠COD ...(vertically opposite angles)
∴ ∠A + ∠AOE = ∠C + ∠COD
⇒ 180° - ∠AEO = 180° - ∠CDO
⇒ ∠AEO = ∠ CDO ….(i)
Now, ∠AEO + ∠OEB = 180° ....(linear pair)
And, ∠CDO + ∠ODB = 180° ....(linear pair)
∴ ∠AEO + ∠OEB = ∠CDO + ∠ODB
⇒ ∠OEB = ∠ODB ....[ Using (i) ]
⇒ ∠CEB = ∠ADB ….(ii)
Now, in ΔABD and ΔCBE,
∠A = ∠C ....(given)
∠ADB = ∠CEB ...[ From (ii) ]
AB = BC ....(given)
⇒ ΔABD ≅ ΔCBE ....(by AAS congruence criterion).
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