a) Denote $\angle BAC = \alpha$ and $\angle ABC = \angle ACB = \beta$.

Fig. 6

Assume that the triangle $BCD$ is isosceles. If $|CB| = |CD|$, then the angles $\angle CBD$, $\angle CDB$ and $\angle BCD$ would be $\frac{\beta}{2}$, $\frac{\beta}{2}$ and $\beta$ respectively, which implies $2 \cdot \frac{\beta}{2} + \beta = 180^\circ$ giving $\beta = 90^\circ$. This is impossible, since the triangle $ABC$ has two angles equal to $\beta$. If $|DB| = |DC|$, then we would get $\frac{\beta}{2} = \beta$, which is also impossible. This leaves the only option $|BC| = |BD|$. Then, the triangles $ABC$ and $BCD$ are similar, since all corresponding angles are the same. Therefore $\angle DBA = \angle DBC = \angle BAC = \angle BAD$, showing that the triangle $ABD$ is isosceles with $|DA| = |DB|$.

b) If the angles of the triangle $ABC$ are $\frac{3}{7} \cdot 180^\circ$, $\frac{2}{7} \cdot 180^\circ$, $\frac{2}{7} \cdot 180^\circ$ (Fig. 7), then $\angle ADB = 180^\circ - \angle BAD - \angle ABD = 180^\circ - \frac{3}{7} \cdot 180^\circ - \frac{1}{7} \cdot 180^\circ = \frac{3}{7} \cdot 180^\circ = \angle BAD$, which shows that the triangle $ABD$ is isosceles with $|BA| = |BD|$. At the same time, the angles in the triangle $BCD$ are $\frac{1}{7} \cdot 180^\circ$, $\frac{2}{7} \cdot 180^\circ$, $\frac{4}{7} \cdot 180^\circ$, which are pairwise different, so the triangle $BCD$ is not isosceles.

Fig. 7