In the given figure, AB = AC and side BA has been produced to D. If AE is the bisector of ∠CAD, prove that AE || BC

From figure,

△ABC is an equilateral triangle.

⇒ ∠BAC = ∠ACB = ∠ABC = 60°

Given,

AE is the bisector of ∠CAD

⇒ ∠CAE = ∠DAE = x (let)

From figure,

⇒ ∠BAC + ∠DAE + ∠CAE = 180° (Linear pair)

⇒ 60° + x + x = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$

⇒ x = 60°

⇒ ∠CAE = ∠DAE = 60°

∴ ∠CAE = ∠ACB = 60°

From figure,

∠CAE and ∠ACB are alternate angles between lines BC and AE and are equal.

Hence, proved that AE || BC.