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Question Number 112626 by Aina Samuel Temidayo last updated on 09/Sep/20

Answered by som(math1967) last updated on 09/Sep/20

a)((x−y)/(b−a))=((y−z)/(c−b))=((z−x)/(a−c))

$$\left.\mathrm{a}\right)\frac{\mathrm{x}−\mathrm{y}}{\mathrm{b}−\mathrm{a}}=\frac{\mathrm{y}−\mathrm{z}}{\mathrm{c}−\mathrm{b}}=\frac{\mathrm{z}−\mathrm{x}}{\mathrm{a}−\mathrm{c}} \\ $$

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

Solution please?

$$\mathrm{Solution}\:\mathrm{please}? \\ $$

Commented by som(math1967) last updated on 09/Sep/20

let (x/(b+c))=(y/(c+a))=(z/(a+b))=k ∴x=k(b+c) y=k(c+a) z=k(a+b) ∴x−y=k(b+c−c−a)=k(b−a) ∴((x−y)/(b−a))=k again y−z=k(c−b) ∴((y−z)/(c−b))=k by same way ((z−x)/(a−c))=k so ((x−y)/(b−a))=((y−z)/(c−b))=((z−x)/(a−c))

$$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{y}}{\mathrm{c}+\mathrm{a}}=\frac{\mathrm{z}}{\mathrm{a}+\mathrm{b}}=\mathrm{k} \\ $$$$\therefore\mathrm{x}=\mathrm{k}\left(\mathrm{b}+\mathrm{c}\right) \\ $$$$\mathrm{y}=\mathrm{k}\left(\mathrm{c}+\mathrm{a}\right) \\ $$$$\mathrm{z}=\mathrm{k}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\therefore\mathrm{x}−\mathrm{y}=\mathrm{k}\left(\mathrm{b}+\mathrm{c}−\mathrm{c}−\mathrm{a}\right)=\mathrm{k}\left(\mathrm{b}−\mathrm{a}\right) \\ $$$$\therefore\frac{\mathrm{x}−\mathrm{y}}{\mathrm{b}−\mathrm{a}}=\mathrm{k} \\ $$$$\mathrm{again}\:\:\mathrm{y}−\mathrm{z}=\mathrm{k}\left(\mathrm{c}−\mathrm{b}\right) \\ $$$$\therefore\frac{\mathrm{y}−\mathrm{z}}{\mathrm{c}−\mathrm{b}}=\mathrm{k} \\ $$$$\mathrm{by}\:\mathrm{same}\:\mathrm{way}\:\frac{\mathrm{z}−\mathrm{x}}{\mathrm{a}−\mathrm{c}}=\mathrm{k} \\ $$$$\mathrm{so}\:\frac{\mathrm{x}−\mathrm{y}}{\mathrm{b}−\mathrm{a}}=\frac{\mathrm{y}−\mathrm{z}}{\mathrm{c}−\mathrm{b}}=\frac{\mathrm{z}−\mathrm{x}}{\mathrm{a}−\mathrm{c}} \\ $$

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

Thanks.

$$\mathrm{Thanks}. \\ $$

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