ΔABC is an isosceles triangle in which AB = AC.Side BA is produced to D such that AD = AB (see Figure). Show that ∠ BCD is a right angle.

In ΔACD,
AB = AC [∵ Given]
∠ACD = ∠D …(1) [∵ Angles opposite to equal sides are equal]
In ΔABC,
AB = AC [∵ Given]
∠B = ∠ACB …(2) [∵ Angles opposite to equal sides are equal]
In ΔDBC,
∠D + ∠B + ∠BCD = 180°
⇒ ∠ACD + ∠ACB + BCD = 180° [∵ From the equation (1) and (2)]
⇒ ∠BCD + ∠BCD = 180° [∵ ∠ACD + ∠ACB = ∠BCD]
⇒ 2∠BCD = 180°
⇒ ∠BCD = (180°/2) = 90°
Hence, ∠BCD is a is a right angle.