ABC is a triangle in which AC = BC and ∠BAC = 50°. Side BC is produced to D such that BC = CD. ∠BAD is equal to :

1. 45°

2. 50°

3. 90°

4. 100°

Given,

AC = BC

∠BAC = ∠ABC = 50°

In △ABC,

By angle sum property of triangle,

∠BAC + ∠ABC + ∠ACB = 180°

⇒ 50° + 50° + ∠ACB = 180°

⇒ 100° + ∠ACB = 180°

⇒ ∠ACB = 180° - 100°

⇒ ∠ACB = 80°

From figure,

∠ACD + ∠ACB = 180° (Linear pair)

⇒ ∠ACD + 80° = 180°

⇒ ∠ACD = 180° - 80°

⇒ ∠ACD = 100°

In △ACD,

AC = CD

∠CAD = ∠ADC = x (let)

By angle sum property of triangle,

⇒ ∠ADC + ∠CAD + ∠ACD = 180°

⇒ x + x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$

⇒ x = 40°.

⇒ ∠CAD = ∠ADC = 40°.

From figure,

∠BAD = ∠BAC + ∠CAD = 50° + 40° = 90°.

Hence, option 3 is the correct option.