Question
If $a: b:: c: d:: e: f$, then prove that $\frac{a e+b f}{a e-b f}=\frac{c e+d f}{c e-d f}$
✓
Answer
$\frac{a}{b}=\frac{c}{d}=\frac{e}{f} $
$\frac{a}{b} \times \frac{e}{f}=\frac{c}{d} \times \frac{e}{f} $
$\Rightarrow \frac{a e}{b f}=\frac{c e}{d f}$
Applying componendo and dividendo
$\frac{a e+b f}{a e-b f}=\frac{c e+d f}{c e-d f}$
Hence, proved.
$\frac{a}{b} \times \frac{e}{f}=\frac{c}{d} \times \frac{e}{f} $
$\Rightarrow \frac{a e}{b f}=\frac{c e}{d f}$
Applying componendo and dividendo
$\frac{a e+b f}{a e-b f}=\frac{c e+d f}{c e-d f}$
Hence, proved.
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