Question

If $\frac{y+z}{a}=\frac{z+x}{b}=\frac{x+y}{c},$ then show that $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ \frac{a}{y+z-x}=\frac{b}{z+x-y}=\frac{c}{x+y-z}=k \\ \mathrm{By}\mathrm{the}\mathrm{theorem}\mathrm{on}\mathrm{equal}\mathrm{ratios},\mathrm{we}\mathrm{get}: \\ k=\frac{b+c}{z+x-y+x+y-z}=\frac{b+c}{2x} \\ \Rightarrow \frac{1}{k}=\frac{2x}{b+c}...\left(\mathrm{i}\right)\left(\mathrm{Invertendo}\right) \\ \mathrm{Now}, \\ k=\frac{c+a}{x+y-z+y+z-x}=\frac{c+a}{2y} \\ \Rightarrow \frac{1}{k}=\frac{2y}{c+a}...\left(\mathrm{ii}\right)\left(\mathrm{Invertendo}\right) \\ \mathrm{Now}, \\ k=\frac{a+b}{y+z-x+z+x-y}=\frac{a+b}{2z} \\ \Rightarrow \frac{1}{k}=\frac{2z}{a+b}...\left(\mathrm{iii}\right)\left(\mathrm{Invertendo}\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\&\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ \frac{2x}{b+c}=\frac{2y}{c+a}=\frac{2z}{a+b} \\ \mathrm{On}\mathrm{multiplying}\mathrm{each}\mathrm{ratio}\mathrm{by}\frac{1}{2},\mathrm{we}\mathrm{get}: \\ \frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b} \end{gathered}$$