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

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

Hence proved.