Question
Given four quantities a, b, c and d are in proportion. Show that
$(a-c) b^2:(b-d) c d=\left(a^2-b^2-a b\right):\left(c^2-d^2-c d\right)$
$(a-c) b^2:(b-d) c d=\left(a^2-b^2-a b\right):\left(c^2-d^2-c d\right)$
✓
Answer
Let $\frac{a}{b}=\frac{c}{d}=k$
$\Rightarrow a=b k$ and $c=d k$
$L . H . S=\frac{(a-c) b^2}{(b-d) c d}$
$=\frac{(b k-d k) b^2}{(b-d) d^2 k} $
$ =\frac{b^2}{d^2}$
R.H.S $=\frac{a^2-b^2-a b}{c^2-d^2-c d}$
$=\frac{b^2 k^2-b^2-b k b}{d^2 k^2-d^2-d k d}$
$=\frac{b^2\left(k^2-1-k\right)}{d^2\left(k^2-1-k\right)}$
$=\frac{b^2}{d^2}$
$\Rightarrow L.H.S = R.H.S$
Hence proved
$\Rightarrow a=b k$ and $c=d k$
$L . H . S=\frac{(a-c) b^2}{(b-d) c d}$
$=\frac{(b k-d k) b^2}{(b-d) d^2 k} $
$ =\frac{b^2}{d^2}$
R.H.S $=\frac{a^2-b^2-a b}{c^2-d^2-c d}$
$=\frac{b^2 k^2-b^2-b k b}{d^2 k^2-d^2-d k d}$
$=\frac{b^2\left(k^2-1-k\right)}{d^2\left(k^2-1-k\right)}$
$=\frac{b^2}{d^2}$
$\Rightarrow L.H.S = R.H.S$
Hence proved
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