Question

If $\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}$ and x + y + z $\ne$ 0 then show that each ratio is $\frac{a+b}{2}.$

Answer 1

$$\begin{gathered} Let: \\ \frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=k \\ \mathrm{By}the\mathrm{theorem}\mathrm{on}\mathrm{equal}\mathrm{ratios},\mathrm{we}\mathrm{get}: \\ k=\frac{ax+by+bx+az+ay+bz}{x+y+x+z+y+z} \\ =\frac{x\left(a+b\right)+y\left(a+b\right)+z\left(a+b\right)}{2\left(x+y+z\right)} \\ =\frac{\left(a+b\right)\left(x+y+z\right)}{2\left(x+y+z\right)} \\ =\frac{a+b}{2} \\ Or, \\ \frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{a+b}{2} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$