In the given figure, △ABC is an equilateral triangle and BC is produced to D such that BC = CD. Prove that AD ⊥ AB.

Given,

△ABC is an equilateral triangle.

⇒ ∠ABC = ∠ACB = ∠BAC = 60°

From figure,

⇒ ∠ACB + ∠ACD = 180° (Linear pair)

⇒ 60° + ∠ACD = 180°

⇒ ∠ACD = 180° - 60°

⇒ ∠ACD = 120°.

In △CAD,

CA = CD (As, BC = CD and BC = CA)

⇒ ∠CAD = ∠CDA = x (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠CDA + ∠ACD = 180°

⇒ x + x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = $\dfrac{60°}{2}$

⇒ x = 30°

⇒ ∠CAD = ∠CDA = 30°

From figure,

⇒ ∠BAD = ∠BAC + ∠CAD = 60° + 30° = 90°.

Hence, proved that AD ⊥ AB.