Δ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Given :

⇒ AB = AC …….(1)

⇒ AD = AB ……(2)

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

⇒ AB = AC = AD

In an isosceles triangle ABC,

⇒ AB = AC

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠ACB = ∠ABC = x (let)

In Δ ACD,

⇒ AC = AD

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠ADC = ∠ACD = y (let)

From figure,

⇒ ∠BCD = ∠ACB + ∠ACD

⇒ ∠BCD = x + y …..(3)

In Δ BCD,

⇒ ∠ABC + ∠BCD + ∠ADC = 180° (Angle sum property of a triangle)

⇒ x + (x + y) + y = 180° [From equation (1), (2) and (3)]

⇒ 2(x + y) = 180°

Substituting value of (x + y) from equation (3) in above equation :

⇒ 2(∠BCD) = 180°

⇒ ∠BCD = $\dfrac{180°}{2}$

⇒ ∠BCD = 90°

Hence, proved that ∠BCD is a right angle.