Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)=k \\ \therefore y+z=\frac{k}{a},z+x=\frac{k}{b}\mathrm{and}x+y=\frac{k}{c} \\ \mathrm{We}\mathrm{have}: \\ x+y-\left(z+x\right)=\frac{k}{c}-\frac{k}{b} \\ \Rightarrow y-z=\frac{k\left(b-c\right)}{bc} \\ \Rightarrow \frac{\left(y-z\right)}{a\left(b-c\right)}=\frac{k}{abc}...\left(1\right) \\ \mathrm{And}, \\ y+z-\left(x+y\right)=\frac{k}{a}-\frac{k}{c} \\ \Rightarrow z-x=\frac{k\left(c-a\right)}{ac} \\ \Rightarrow \frac{z-x}{b\left(c-a\right)}=\frac{k}{abc}...\left(2\right) \\ \mathrm{And}, \\ z+x-\left(y+z\right)=\frac{k}{b}-\frac{k}{a} \\ \Rightarrow x-y=\frac{k\left(a-b\right)}{ba} \\ \Rightarrow \frac{x-y}{c\left(a-b\right)}=\frac{k}{abc}...\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}: \\ \frac{\left(y-z\right)}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}=\frac{k}{abc} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$