jmc

Since $AB=AC$, triangle $ABC$ must be an isosceles triangle and the measures of $\angle ABC$ and $\angle ACB$ must be equal. Continuing, since $\overline{BD}$ bisects angle $ABC$, we have that the measures of $\angle ABD$ and $\angle BDC$ are equal. Finally, since $BD=BC$, triangle $BDC$ must also be an isosceles triangle so the measures of $\angle BDC=\angle BCD$. Now if we consider triangle $BDC$, we know that angles $BDC$ and $BCD$ have equal angle measures and angle $DBC$ has an angle measure that is half that of the other two. Since these three angle measures must add up to $180^{\circ }$, we have that $\angle DBC$ has measure $36^{\circ }$ and angles $BDC$ and $BCD$ have measures $72^{\circ }$.

Now, since $\angle ABC\cong \angle ACB$ and $\angle ACB$ has measure $72^{\circ }$, we know that $\angle A$ must have an angle measure of $180-72-72=\boxed{36}$ degrees.