Question

If in a triangle $ABC,\angle C={60}^{\circ },$ then prove that $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$

Answer 1

Formula:

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

Given, $\angle C={60}^{\circ }$

$\therefore c^2=a^{+}{b}^2-2ab\cos 60$

$c^2=a^{+}{b}^2-ab$..............(1)

To prove:

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

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

$\frac{a+b+2c}{(a+c)(b+c)}=\frac{3}{a+b+c}$

$(a+b+2c)(a+b+c)=3\left[\right(a+c\left)\right(b+c\left)\right]$

$a^2+ab+ac+ba+b^2+bc+2ca+2cb+2c^2=3ab+3ac+3cb+3c^2$

$a^2+b^2+2ab-3ab=3c^2-2c^2$

$a^2+b^2-ab=c^2$.............(2)

From (1) and (2) it's clear that,

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