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 =
⇒ ∠BCD = 90°
Hence, proved that ∠BCD is a right angle.