In a ΔABC, AB = AC and BA is produced to D such that AC = AD. Then the ∠BCD is

Given : AB = AC = AD

To find : $$\angle$$BCD = ?

Solution : Since, AB = AC, => $$\angle$$ABC = $$\angle$$ACB

and $$\angle$$ACD = $$\angle$$ADC

and $$\angle$$BCD = $$\angle$$ACB + $$\angle$$ACD

In $$\triangle$$BCD

=> $$\angle$$CDB + $$\angle$$DBC + $$\angle$$BCD = 180°

=> $$\angle$$CDB + $$\angle$$DBC + ($$\angle$$ACD + $$\angle$$ACB) = 180°

=> 2$$\angle$$ACB + 2$$\angle$$ACD = 180°

=> $$\angle$$ACB + $$\angle$$ACD = 90°

=> $$\angle$$BCD = 90°