Question
If $a : b = c : d$, show that $(a - c) b^2 : (b - d) cd = (a^2 - b^2 - ab) : (c^2 - d^2 - cd)$.
✓
Answer
Let $\frac{a}{b}=\frac{c}{d}=k$
$\Rightarrow a = bk$ and $c = dk$
L.H.S.
$=\frac{(a-c) b^2}{(b-d) c d}=\frac{(b k-d k) b^2}{(b-d) d k \cdot d}$
$=\frac{b^2 k(b-d)}{d 2 k(b-d)}=\frac{b^2}{d^2}$
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 \cdot b}{d^2 k^2-d^2-d k \cdot d} $
$=\frac{b^2\left(k^2-k-1\right)}{d^2\left(k^2-k-1\right)} $
$=\frac{b^2}{d^2}$
L.H.S. = R.H.S.
Hence proved.
$\Rightarrow a = bk$ and $c = dk$
L.H.S.
$=\frac{(a-c) b^2}{(b-d) c d}=\frac{(b k-d k) b^2}{(b-d) d k \cdot d}$
$=\frac{b^2 k(b-d)}{d 2 k(b-d)}=\frac{b^2}{d^2}$
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 \cdot b}{d^2 k^2-d^2-d k \cdot d} $
$=\frac{b^2\left(k^2-k-1\right)}{d^2\left(k^2-k-1\right)} $
$=\frac{b^2}{d^2}$
L.H.S. = R.H.S.
Hence proved.
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