2. Given an isosceles triangle $ABC, \overline{AB}=\overline{AC}$. The angle bisector of $\angle ABC$ intersects side $AC$ at point $D$. If $\overline{BC}=\overline{AB}+2 \overline{AD}$, calculate the measures of the angles of the triangle.
Solution. Let $\overline{B C}=a, \overline{A B}=\overline{A C}=b$ and $\overline{A D}=p$. From the conditions in the problem, we obtain the system

$$
\left\{\begin{array}{l}
a=b+2 p \\
\frac{a}{b}=\frac{b-p}{p}
\end{array}\right.
$$

(the second equality holds because $A D$ is the angle bisector of $\angle A B C$). Then

$$
b+2 p=\frac{b(b-p)}{p}, b^{2}-2 b p=2 p^{2}
$$

which simplifies to $(b-p)^{2}=3 p^{2}$. Thus, $b=p(1+\sqrt{3})$ and $a=p(3+\sqrt{3})$. Since

$$
\cos \measuredangle A B C=\frac{a}{2 b}=\frac{3+\sqrt{3}}{2(1+\sqrt{3})}=\frac{(3+\sqrt{3})(\sqrt{3}-1)}{2(3-1)}=\frac{\sqrt{3}}{2},
$$

it follows that $\measuredangle A B C=\measuredangle A C B=30^{\circ}$ and $\measuredangle C A B=120^{\circ}$.