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Question Number 211651 by MATHEMATICSAM last updated on 15/Sep/24

$$\mathrm{In}\mathrm{triangle}\mathrm{ABC},\mathrm{\angle }\mathrm{C}=60°.\mathrm{If}\mathrm{the}\mathrm{length}$$$$\mathrm{of}\mathrm{opposite}\mathrm{sides}\mathrm{of}\mathrm{\angle }\mathrm{A},\mathrm{\angle }\mathrm{B}\mathrm{and}\mathrm{\angle }\mathrm{C}\mathrm{are}$$$$a,b\mathrm{and}c\mathrm{respectively}\mathrm{then}\mathrm{prove}\mathrm{that}$$$$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}.$$

Answered by som(math1967) last updated on 15/Sep/24

$$\frac{a^2+b^2-c^2}{2ab}=cos60$$$$\Rightarrow a^2+b^2-c^2=\frac{1}{2}\times 2ab$$$$\Rightarrow a^2+b^2=ab+c^2$$$$\Rightarrow a^2+b^2+bc+ca=ab+bc+ca+c^2$$$$\Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)$$$$\Rightarrow \frac{b}{a+c}+\frac{a}{b+c}=1$$$$\Rightarrow \frac{b}{a+c}+1+\frac{a}{b+c}+1=1+2$$$$\Rightarrow \frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$$$$\Rightarrow \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$$

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