Question
If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.
✓
Answer
$
\begin{aligned}
& \frac{a+3 b+2 c+6 d}{a-3 b-2 c+6 d}=\frac{a+3 b-2 c-6 d}{a-3 b+2 c-6 d} \\
& \Rightarrow \frac{a+3 b+2 c+6 d}{a+3 b-2 c-6 d}=\frac{a-3 b+2 c-6 d}{a-3 b-2 c+6 d} \ldots \text { (by altenendo) }
\end{aligned}
$Applying componendo and dividendo
$
\begin{aligned}
& \frac{a+3 b+2 c+6 d+a+3 b-2 c-6 d}{a+3 b+2 c+6 d-a-3 b+2 c+6 d} \\
& =\frac{a-3 b+2 c-6 d+a-3 b-2 c+6 d}{a-3 b+2 c-6 d-a+3 b+2 c-6 d} \\
& \Rightarrow \frac{2(a+3 b)}{2(2 c+6 d)}=\frac{2(a-3 b)}{2(2 c-6 d)} \\
& \Rightarrow \frac{a+3 b}{2 c+6 d}=\frac{a-3 b}{2 c-6 d} \ldots \text { (Dividing by 2) } \\
& \Rightarrow \frac{a+3 b}{a-3 b}=\frac{2 c+6 d}{2 c-6 d} \ldots \text { (By alternendo) }
\end{aligned}
$Again applying componendo and dividendo
$
\frac{a+3 b+a-3 b}{a+3 b-a+3 b}=\frac{2 c+6 d+2 c 6 d}{2 c+6 d-2 c+6 d}
$
$\begin{aligned} & \Rightarrow \frac{2 a}{6 b}=\frac{4 c}{12 d}=\frac{2 c}{6 d} \\ & \Rightarrow \frac{a}{b}=\frac{c}{d} \cdot \quad \ldots\left[\text { Dividing by } \frac{2}{6}\right]\end{aligned}$
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