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

∠ ABD = ∠ ACD ∠ ABD > ∠ ACD ∠ ABD < ∠ACD None of the above

∠ ABD = ∠ ACD

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