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Question Number 214402 by golsendro last updated on 07/Dec/24

$$\mathrm{If}\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\mathrm{then}\mathrm{find}$$$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}.$$

Answered by efronzo1 last updated on 07/Dec/24

$$\mathrm{Let}\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}=\mathrm{m}$$$$\mathrm{then}\mathrm{a}+\mathrm{b}=\mathrm{mc};\mathrm{b}+\mathrm{c}=\mathrm{ma};\mathrm{a}+\mathrm{c}=\mathrm{mb}$$$$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{b}+\mathrm{c}+\mathrm{a}+\mathrm{c}=\mathrm{mc}+\mathrm{ma}+\mathrm{mb}$$$$2\mathrm{a}+2\mathrm{b}+2\mathrm{c}=\mathrm{m}(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$2(\mathrm{a}+\mathrm{b}+\mathrm{c})=\mathrm{m}(\mathrm{a}+\mathrm{b}+\mathrm{c})$$$$\mathrm{m}=2\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}=2$$

Answered by Rasheed.Sindhi last updated on 07/Dec/24

$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}$$$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}+1=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}+1=\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}+1$$$$\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}$$$$a+b+c=0\mid \frac{1}{c}=\frac{1}{a}=\frac{1}{b}$$$$\{\begin{array}{l}a+b+c=0\Rightarrow a+b=-c\Rightarrow \frac{a+b}{c}=\frac{-c}{c}=-1\\ a=b=c\Rightarrow \frac{a+b}{c}=\frac{a+a}{a}=\frac{2a}{a}=2\end{array}$$$$\frac{a+b}{c}=-1or2$$

Answered by Rasheed.Sindhi last updated on 07/Dec/24

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

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