In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD. Prove that : ∠BCD = 90°.

In △ ABC,

⇒ AB = AC (Given)

⇒ ∠B = ∠C (Angles opposite to equal sides are equal) ………(1)

In △ ACD,

⇒ AC = AD (Given)

⇒ ∠ADC = ∠ACD (Angles opposite to equal sides are equal) …….(2)

Adding equation (1) and (2), we get :

⇒ ∠B + ∠ADC = ∠C + ∠ACD

⇒ ∠B + ∠ADC = ∠BCD ….(3)

In △ BCD,

⇒ ∠B + ∠ADC + ∠BCD = 180° (By angle sum property of triangle)

⇒ ∠BCD + ∠BCD = 180°

⇒ 2∠BCD = 180°

⇒ ∠BCD = $\dfrac{180°}{2}$ = 90°.

Hence, proved that ∠BCD = 90°.