Question
If $\frac{a}{b}=\frac{c}{d}$ show that $(a+b):(c+d)=\sqrt{a^2+b^2}: \sqrt{c^2+d^2}$
✓
Answer
Let $\frac{a}{b}=\frac{c}{d}=k$ (say)
$`\Rightarrow a = bk , c = dk$
$\text { L.H.S }=\frac{a+b}{c+d}$
$=\frac{b k+b}{d k+d} $
$ =\frac{b(k+1)}{d(k+1)} $
$=\frac{b}{d}$
$\text { R.H.S }=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} $
$ =\frac{\sqrt{(b k)^2+b^2}}{\sqrt{(d k)^2+d^2}} $
$=\frac{\sqrt{b^2\left(k^2+1\right)}}{\sqrt{d^2\left(k^2+1\right)}}$
$ =\frac{\sqrt{b^2}}{\sqrt{d^2}} $
$=\text { b/d } $
$ \therefore \text { L.H.S }=\text { R.H.S }$
$`\Rightarrow a = bk , c = dk$
$\text { L.H.S }=\frac{a+b}{c+d}$
$=\frac{b k+b}{d k+d} $
$ =\frac{b(k+1)}{d(k+1)} $
$=\frac{b}{d}$
$\text { R.H.S }=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} $
$ =\frac{\sqrt{(b k)^2+b^2}}{\sqrt{(d k)^2+d^2}} $
$=\frac{\sqrt{b^2\left(k^2+1\right)}}{\sqrt{d^2\left(k^2+1\right)}}$
$ =\frac{\sqrt{b^2}}{\sqrt{d^2}} $
$=\text { b/d } $
$ \therefore \text { L.H.S }=\text { R.H.S }$
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