In the given figure, AD = AE and ∠BAD = ∠CAE. Prove that : AB = AC.

In △ADE,

AD = AE

⇒ ∠ADE = ∠AED (Angles opposite to equal sides in atriangle are equal)

From figure,

⇒ ∠ADE + ∠ADB = 180° (Linear pair)

⇒ ∠ADB = 180° - ∠ADE ….(1)

⇒ ∠AED + ∠AEC = 180° (Linear pair)

⇒ ∠AEC = 180° - ∠AED

⇒ ∠AEC = 180° - ∠ADE ….(2) (∵ ∠ADE = ∠AED)

From eq.(1) and (2), we have:

⇒ ∠AEC = ∠ADB

In △ABD and △ACE,

⇒ AD = AE (Given)

⇒ ∠ADB = ∠AEC (Proved above)

⇒ ∠BAD = ∠CAE (Given)

∴ △ABD ≅ △ACE (By A.S.A axiom)

⇒ AB = AC (Corresponding parts of congruent triangles are equal)

Hence, proved that AB = AC.