Question
If $a: b=c: d$, then prove that $\frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}$
✓
Answer
$\frac{ a }{ b }=\frac{ c }{ d } \Rightarrow a =\frac{ bc }{ d }$
LHS
$\frac{a^2+a b+b^2}{a^2-a b+b^2} $
$=\frac{\left(\frac{b c}{d}\right)^2+\left(\frac{b c}{d}\right) b+b^2}{\left(\frac{b c}{d}\right)^2-\left(\frac{b c}{d}\right) b+b^2} $
$=\frac{b^2 c^2+b^2 c d+d^2 b^2}{b^2 c^2-b^2 c d+d^2 b^2} $
$=\frac{b^2\left(c^2+c d+d\right)}{b^2\left(c^2-c d+d^2\right)}=\frac{c^2+c d+d^2}{c^2-c d+d^2}=\text { RHS }$
LHS $=$ RHS
Hence, proved.
LHS
$\frac{a^2+a b+b^2}{a^2-a b+b^2} $
$=\frac{\left(\frac{b c}{d}\right)^2+\left(\frac{b c}{d}\right) b+b^2}{\left(\frac{b c}{d}\right)^2-\left(\frac{b c}{d}\right) b+b^2} $
$=\frac{b^2 c^2+b^2 c d+d^2 b^2}{b^2 c^2-b^2 c d+d^2 b^2} $
$=\frac{b^2\left(c^2+c d+d\right)}{b^2\left(c^2-c d+d^2\right)}=\frac{c^2+c d+d^2}{c^2-c d+d^2}=\text { RHS }$
LHS $=$ RHS
Hence, proved.
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