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

Question

How to solve this $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$. Given that $\angle C=60°$. Please answer as soon as possible. Tomorrow is my test

Answer 1

$$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\iff$$ $$(a+b+c)(a+b+2c)=3(a+c)(b+c)\iff$$ $$c^2=a^2+b^2-ab=a^2+b^2-2ab\cos(60^o)$$

Answer 2

the left side of your equation is given by $$\left( \sin \left( \alpha \right) +1/2\,\sqrt {3} \right) ^{-1}+ \left( \sin \left( \pi/3+\alpha \right) +1/2\,\sqrt {3} \right) ^{-1} $$ and the right side $$3\, \left( \sin \left( \alpha \right) +\sin \left( \pi/3+\alpha \right) +1/2\,\sqrt {3} \right) ^{-1} $$ and the Difference is $$4\, \left( \sin \left( \alpha \right) \right) ^{2}-4\,\sin \left( \pi /3+\alpha \right) \sin \left( \alpha \right) -3+4\, \left( \sin \left( \pi/3+\alpha \right) \right) ^{2} $$ simplifying this we obtain $$4\sin(\alpha)^2-4\sin((1/3)\pi+\alpha)\sin(\alpha)-3+4\sin((1/3)\pi+\alpha)^2$$ simplifying this we get $$3\, \left( \sin \left( \alpha \right) \right) ^{2}-3+3\, \left( \cos \left( \alpha \right) \right) ^{2} $$ and this is Zero.