If in a $\triangle A B C, \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$, then $\angle C$ is equal to
If in a $\triangle A B C, \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$, then $\angle C$ is equal to
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
Solution
In $\triangle A B C$
$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$
Let $\angle C=60^{\circ}$, then
$\cos C=\frac{\pi}{3}$
$\begin{aligned}
& \Rightarrow a^2+b^2-c^2=a b \\
& \Rightarrow b^2+b c+a^2+a c=a b+a c+b c+c^2 \\
& \Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)
\end{aligned}$
Divide by $(a+c)(b+c)$ and add 2 on both sides, we get
$\begin{aligned}
& 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\
\Rightarrow \quad & \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}
\end{aligned}$
So, $\angle C$ should be $60^{\circ}$.