Question

If $a,b,c,d$ are in continuous proportion, then

$a-c:b-d=\sqrt{a^2+c^2}:\sqrt{b^2+d^2}$ is true.

Answer 1

Now $a,b,c,d$ are in continuous proportion. So,

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

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

Given to prove,

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

Squaring both sides,

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

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

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

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

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

Putting the values we have,

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

which is true.