In the given figure, △ABC is an equilateral triangle whose base BC is produced to D such that BC = CD. Calculate :

(i) ∠ACD

(ii) ∠ADC

(i) Given, △ABC is an equilateral triangle.

∠BAC = ∠ACB = ∠ABC = 60°

From figure,

⇒ ∠ACB + ∠ACD = 180° (Linear pair)

⇒ 60° + ∠ACD = 180°

⇒ ∠ACD = 180° - 60°

⇒ ∠ACD = 120°.

Hence, ∠ACD = 120°.

(ii) In △ACD,

AC = CD

⇒ ∠CAD = ∠ADC = x (let) (Angles opposite to equal sides in a triangle are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠ADC + ∠ACD = 180°

⇒ x + x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = $\dfrac{60°}{2}$

⇒ x = 30°

⇒ ∠CAD = ∠ADC = 30°.

Hence, ∠ADC = 30°.