Task A-3.5. (4 points)

If the lengths of the sides of a triangle satisfy the equation $\frac{a+b}{b+c}=\frac{c}{a-b}$, determine the measure of the largest angle of this triangle.
## First solution.

From the given equality, we have in sequence

$$
\begin{gathered}
(a-b)(a+b)=c(b+c) \\
a^{2}-b^{2}=b c+c^{2} \\
b^{2}+c^{2}-a^{2}=-b c
\end{gathered}
$$

which means $\frac{b^{2}+c^{2}-a^{2}}{2 b c}=-\frac{1}{2}$.

Therefore, $\cos \alpha=-\frac{1}{2}$

and finally $\alpha=120^{\circ}$, which is clearly the largest angle in the triangle.