Question

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ then show that

$(b^2+d^2+f^2)\times (a^2+c^2+e^2)=(ab+cd+ef{)}^2$

• A. True
• B. False

Answer 1

The correct option is A True

Given $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$

We can write above equation as follows

$⟹\frac{ab}{b^2}=\frac{cd}{d^2}=\frac{ef}{f^2}=k---eqn\left(1\right)$

Similarly $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$

$⟹\frac{a^2}{ab}=\frac{c^2}{cd}=\frac{e^2}{ef}=k---eqn\left(2\right)$

So by theorem on equal ratios we can write $eqn\left(1\right)$ as

$\frac{ab}{b^2}=\frac{cd}{d^2}=\frac{ef}{f^2}=\frac{ab+cd+ef}{b^2+d^2+e^2}=k---eqn\left(3\right)$

SImilar;y from $eqn\left(2\right)$ we get

$\frac{a^2}{ab}=\frac{c^2}{cd}=\frac{e^2}{ef}=\frac{a^2+c^2+e^2}{ab+cd+ef}=k---eqn\left(4\right)$

So from $eqn\left(3\right)$ and $eqn\left(4\right)$

$\frac{ab+cd+ef}{b^2+d^2+f^2}=\frac{a^2+c^2+e^2}{ab+cd+ef}$

So after cross-multiplication

$(b^2+d^2+f^2)\times (a^2+c^2+e^2)=(ab+cd+ef{)}^2$ $\left[henceproved\right]$