In triangle \(ABC\) , \(AB=AC\) and \(D\) is a point on \(\overline{AC}\) so that \(\overline{BD}\) bisects angle \(ABC\). If \(BD=BC\), what is the measure, in degrees, of angle \(A\)?
Since AB = AC, then
Angle ABC = angle ACB
And since BD = BC
Then Angle BDC = Angle DCB
But DCB = ACB
So....triangles ABC and BDC are similar
Call angle ABC = 2theta
But since ABC is bisected, then DBC is also = theta
And angle DBC = angle BAC = theta
Then...in triangle ABC
Angle ABC + Angle ACB + Angle BAC = 180
2theta + 2theta + theta = 180
5theta = 180
theta = 36°= BAC = "Angle A "