If ABC and DBC are two isosceles triangles on the same base BC. Then:

In ΔABC and ΔDBC,

∠ABC = ∠ACB, and ∠DBC = ∠DCB

Let ∠ABC and ∠ACB = xº and ∠DBC and ∠DCB = yº

∠ABD = xº - yº and ∠ACD = xº - yº

So, we can say that ∠ABD = ∠ACD i.e. x - yº