Question

If b² = ac then show that $\frac{a-c}{b+c}=\frac{a-b}{b}.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {b}^2=ac\Rightarrow \frac{a}{b}=\frac{b}{c} \\ \mathrm{Let}: \\ \frac{a}{b}=\frac{b}{c}=k \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ a=bk\mathrm{and}b=ck \\ \therefore a=c{k}^2 \\ \mathrm{Now}, \\ \mathrm{LHS}=\frac{a-c}{b+c} \\ =\frac{c{k}^2-c}{ck+c} \\ =\frac{c\left(k^2-1\right)}{c\left(k+1\right)} \\ =\frac{\left(k-1\right)\left(k+1\right)}{k+1} \\ =k-1 \\ \mathrm{And}, \\ \mathrm{RHS}=\frac{a-b}{b} \\ =\frac{c{k}^2-ck}{ck} \\ =\frac{ck\left(k-1\right)}{ck} \\ =k-1 \\ \mathrm{LHS}=\mathrm{RHS} \\ \therefore \frac{a-c}{b+c}=\frac{a-b}{b} \end{gathered}$$

Hence proved.