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Triangle with 60 degree show
I do not understand how to do the following problem. I have attempted to use cos rule, sin rule and even constructions to make use of the 60 degree information but i still cant reach the answer.
By the Law of Cosines
$$c^2=a^2+b^2ab$$
hence
\begin{align*}
&\frac{3}{a+b+c}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\[4pt]
=\;&\frac{3(a+c)(b+c)(a+b+c)(b+c)(a+b+c)(a+c)}{(a+b+c)(a+c)(b+c)}\\[4pt]
=\;&\frac{(3ab+3ac+3bc+3c^2)(ab+ac+2bc+b^2+c^2)(ab+bc+2ac+a^2+c^2)}{(a+b+c)(a+c)(b+c)}
\\[4pt]
=\;&\frac{c^2(a^2+b^2ab)}{(a+b+c)(a+c)(b+c)}\\[4pt]
=\;&\;0
\\[4pt]
\end{align*}
As to why the simplification above was destined to work, the idea is that the Law of Cosines, applied to the angle $C$, yields an ifandonlyif condition for $C=60^{\circ}$, hence, if $C=60^{\circ}$ implies some other identity relating $a,b,c$, we would expect it to be an algebraic consequence of the Law of Cosines.

By the Law of Cosines
$$c^2=a^2+b^2ab$$
hence
\begin{align*}
&\frac{3}{a+b+c}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\[4pt]
=\;&\frac{3(a+c)(b+c)(a+b+c)(b+c)(a+b+c)(a+c)}{(a+b+c)(a+c)(b+c)}\\[4pt]
=\;&\frac{(3ab+3ac+3bc+3c^2)(ab+ac+2bc+b^2+c^2)(ab+bc+2ac+a^2+c^2)}{(a+b+c)(a+c)(b+c)}
\\[4pt]
=\;&\frac{c^2(a^2+b^2ab)}{(a+b+c)(a+c)(b+c)}\\[4pt]
=\;&\;0
\\[4pt]
\end{align*}
As to why the simplification above was destined to work, the idea is that the Law of Cosines, applied to the angle $C$, yields an ifandonlyif condition for $C=60^{\circ}$, hence, if $C=60^{\circ}$ implies some other identity relating $a,b,c$, we would expect it to be an algebraic consequence of the Law of Cosines.
20180514 08:49:32