If $a : b = c : d$, prove that $\left(abcd\right) \left(a^{-2} + b^{-2} + c^{-2} + d^{-2}\right) = a^2 + b^2 + c^2 + d^2$
Given: $\;$ $a : b = c : d$
$\implies$ $ad = bc$ $\;\;\; \cdots \; (1)$
Now,
$\begin{aligned}
LHS & = \left(abcd\right) \left(a^{-2} + b^{-2} + c^{-2} + d^{-2}\right) \\\\
& = \left(abcd\right) \left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2}\right) \\\\
& = \left(abcd\right) \left(\dfrac{b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2}{a^2 b^2 c^2 d^2}\right) \\\\
& = \dfrac{b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2}{abcd} \\\\
& = \dfrac{bcd}{a} + \dfrac{acd}{b} + \dfrac{abd}{c} + \dfrac{abc}{d} \\\\
& = bc \left(\dfrac{a}{d} + \dfrac{d}{a}\right) + ad \left(\dfrac{c}{b} + \dfrac{b}{c}\right) \\\\
& = ad \left(\dfrac{a}{d} + \dfrac{d}{a}\right) + bc \left(\dfrac{c}{b} + \dfrac{b}{c}\right) \;\;\; \left[\text{in view of equation (1)}\right] \\\\
& = a^2 + d^2 + c^2 + b^2 \\\\
& = RHS
\end{aligned}$
Hence proved.