Question
If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, prove that$(a b+c d+e f)^2=\left(a^2+c^2+e^2\right)\left(b^2+d^2+f^2\right)$.
✓
Answer
Let $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$ then
$a = bk , c = dk$ and $e = fk$
L.H.S.
$=(a b+c d+e f)^2 $
$=(b k \cdot b+d k \cdot d+f k \cdot f)^2 $
$=k^2\left(b^2+d^2+f^2\right)^2$
R.H.S.
$=\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)\left(b^2+d^2+f^2\right) $
$=k^2\left(b^2+d^2+f^2\right)\left(b^2+d^2+f^2\right) $
$=k^2\left(b^2+d^2+f^2\right)^2$
L.H.S. = R.H.S.
Hence proved.
$a = bk , c = dk$ and $e = fk$
L.H.S.
$=(a b+c d+e f)^2 $
$=(b k \cdot b+d k \cdot d+f k \cdot f)^2 $
$=k^2\left(b^2+d^2+f^2\right)^2$
R.H.S.
$=\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)\left(b^2+d^2+f^2\right) $
$=k^2\left(b^2+d^2+f^2\right)\left(b^2+d^2+f^2\right) $
$=k^2\left(b^2+d^2+f^2\right)^2$
L.H.S. = R.H.S.
Hence proved.
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