It is given in the question that,

In ∆ABC,

AB = AC

We know that, angles opposite to equal sides are equal

∴ ∠ACB = ∠ABC

Now, in we have:

AC = AD

∠ADC = ∠ACD (The Angles opposite to equal sides are equal)

By using angle sum property in triangle BCD, we get:

∠ABC + ∠BCD + ∠ADC = 180^o

∠ACB + ∠ACB + ∠ACD + ∠ACD = 180^o

2 (∠ACB + ∠ACD) = 180^o

2 (∠BCD) = 180^o

∠BCD =

∠BCD = 90^o

Hence, proved