Question

If in a triangle ABC, $\angle C={60}^0$, then prove that $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$.

Answer 1

Since $\angle ={60}^o$, we have
$c^2=a^2+b^2-2ab\cos C$
$=a^2+b^2-2ab\cos {60}^o$
or $c^2=a^2+b^2-ab$ ...........(1)
Also $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$
if $\frac{a+b+2c}{(a+c)(b+c)}=\frac{3}{a+b+c}$
i.e. if $(a+b+2c)(a+b+c)=3(a+c)(b+c)$
i.e. if $(a+b{)}^2+2c^2+3c(a+b)=3(ab+ac+bc+c^2)$
i.e. if $a^2+b^2+2ab+2c^2+3ca+3cb=3ab+3ac+3bc+3c^2$
i.e. if $a^2+b^2-ab=c^2$
which is the same as (1).