Question

If $a,b,c,d$ are in continued proportion, prove that

$\sqrt{a^2+c^2}:\sqrt{b^2+d^2}=\sqrt{ac+\frac{c^3}{a}}:\sqrt{bd+\frac{d^3}{b}}$.

Answer 1

Now here $a,b,c,d$ are in continuous proportion, so

$\frac{a}{b}=\frac{c}{d}\to \frac{a}{c}=\frac{b}{d}=k$

let it be $k$ where $k$ is a constant.

Given to prove,

$\frac{\sqrt{a^2+c^2}}{\sqrt{b^2+d^2}}=\frac{\sqrt{ac+\frac{c^3}{a}}}{\sqrt{bd+\frac{d^3}{b}}}$

Squaring both sides,

$\frac{a^2+c^2}{b^2+d^2}=\frac{ac+\frac{c^3}{a}}{bd+\frac{d^3}{b}}$

$\frac{a^2+c^2}{ac+\frac{c^3}{a}}=\frac{b^2+d^2}{bd+\frac{d^3}{b}}$

$\frac{\frac{a}{c}+\frac{c}{a}}{1+\frac{c^2}{a^2}}=\frac{\frac{b}{d}+\frac{d}{b}}{1+\frac{d^2}{b^2}}$

Putting the values,

$\frac{k+\frac{1}{k}}{1+\frac{1}{k^2}}=\frac{k+\frac{1}{k}}{1+\frac{1}{k^2}}$

which is true.