Question
If $\frac{a}{c}=\frac{c}{d}=\frac{c}{f}$ prove that $:\left(b^2+d^2+f^2\right)\left(a^2+c^2+e^2\right)=(a b+$ $c d+e f)^2$

Answer


$\begin{aligned} & \frac{a}{c}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\ & \therefore a=b k_t c=d k^2 e=f k \\ & \text { L.H.S. }=\left(b^2+d^2+f^2\right)\left(a^2+c^2+e^2\right) \\ & =\left(b^2+d^2+f^2\right)\left(b^2 k^2+d^2 k^2+f^2 k^2\right) \\ & =k^2\left(b^2+d^2+f^2\right) k^2\left(b^2+d^2+f^2\right) \\ & =k^2\left(b^2+d^2+f^2\right)^2 \\ & \text { R.H.S. }=(a b+c d+e f)^2 \\ & =(b . k b+d k \cdot d+f k . f)^2 \\ & =\left(k b^2+k d^2+k f^2\right) \\ & =k^2\left(b^2+d^2+f^2\right)^2 \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$

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