Question
If $a: b=c: d$, then prove that $\frac{a^2+c^2}{b^2+d^2}=\frac{a c}{b c}$
✓
Answer
$\frac{ a }{ b }=\frac{ c }{ d } $
$\Rightarrow a =\frac{ bc }{ d }$
To prove,
$\frac{ a ^2+ c ^2}{ b ^2+ d ^2}=\frac{ ac }{ bd }$
$\ce{LHS}$
$\frac{a^2+c^2}{b^2+d^2}$
$=\frac{\left(\frac{b c}{d}\right)^2+c^2}{b^2+d^2}$
$=\frac{\frac{b^2 c^2}{d^2}+c^2}{b^2+d^2}$
$=\frac{c^2\left(b^2+d^2\right)}{d^2\left(b^2+d^2\right)}$
$=\frac{c^2}{d^2}$
$\text { RHS }$
$\frac{ ac }{ bd }$
$=\frac{\frac{ bc }{ d c}}{ bd }$
$=\frac{ bc ^2}{ bd ^2}$
$=\frac{ c ^2}{ d ^2}$
$\text { LHS }=\text { RHS }$
$\Rightarrow a =\frac{ bc }{ d }$
To prove,
$\frac{ a ^2+ c ^2}{ b ^2+ d ^2}=\frac{ ac }{ bd }$
$\ce{LHS}$
$\frac{a^2+c^2}{b^2+d^2}$
$=\frac{\left(\frac{b c}{d}\right)^2+c^2}{b^2+d^2}$
$=\frac{\frac{b^2 c^2}{d^2}+c^2}{b^2+d^2}$
$=\frac{c^2\left(b^2+d^2\right)}{d^2\left(b^2+d^2\right)}$
$=\frac{c^2}{d^2}$
$\text { RHS }$
$\frac{ ac }{ bd }$
$=\frac{\frac{ bc }{ d c}}{ bd }$
$=\frac{ bc ^2}{ bd ^2}$
$=\frac{ c ^2}{ d ^2}$
$\text { LHS }=\text { RHS }$
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