Question

In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 

Answer 1

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).