ΔABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that ∠BCD = 90°.

In ΔABC, AB = AC
           ∴ ഋ = ∠C= ∠4                              ......(i) 
  Since, AB = AC ...... (Given)
     and AB = AD ...... (Produced)
 ∴ AD = AC 
Now, in ΔACD, AD = AC
∴ ∠D = ∠C = ∠3                                        ...... (ii) 
Adding eqn (i) and eqn (ii), we get 
   ∠B + ∠D = ∠4 + ∠3
⇒ ∠B + ∠D = ∠BCD
Now in ΔBCD, we have
     ∠B + ∠BCD + ∠D = 180°
⇒ (∠B + ∠D) + ∠BCD = 180°
      ⇒ ∠BCD + ∠BCD = 180°
        ⇒ 2 ∠BCD = 180°
 ...  󐯍 =   = 90°