3. Calculate the angle $\gamma$ in triangle $ABC$, if

$$
\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}
$$

where $a, b, c$ are the lengths of the sides of triangle $ABC$ opposite to vertices $A, B, C$ respectively, and $\gamma$ is the angle at vertex $C$.
Solution. We multiply the equation $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$ by $a+b+c>0$ and obtain the equivalent equation

$$
\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3
$$

or

$$
\frac{b}{a+c}+\frac{a}{b+c}=1
$$

We multiply the last equation by $(a+c)(b+c)>0$ and obtain the equivalent equation

$$
b^{2}+b c+a^{2}+a c=a b+a c+c b+c^{2}
$$

or

$$
c^{2}=b^{2}+a^{2}-a b
$$

On the other hand, according to the cosine theorem, we have

$$
c^{2}=a^{2}+b^{2}-2 a b \cos \gamma
$$

and due to (*) we obtain $\cos \gamma=\frac{1}{2}$, from which we get $\gamma=60^{\circ}$.