Question

If $$\frac{a+3b+2c+6d}{a-3b+2c-6d} = \frac{a+3b-2c-6d}{a-3b-2c+6d}$$, prove that $$\frac{a}{b} = \frac{c}{d}$$.

\[ \begin{array}{l} \left[ \textbf{Hint :}\ \text{We have}\ \dfrac{(a + 3b) + (2c + 6d)}{(a + 3b) - (2c + 6d)} = \dfrac{(a - 3b) + (2c - 6d)}{(a - 3b) - (2c - 6d)} \right. \\[16pt] \text{Now apply componendo and dividendo.}] \end{array} \]

Answer 1

Given: $$\frac{a + 3b + 2c + 6d}{a - 3b + 2c - 6d} = \frac{a + 3b - 2c - 6d}{a - 3b - 2c + 6d}$$

To prove: $$\frac{a}{b} = \frac{c}{d}$$

Proof:

1. $$\frac{(a + 3b) + (2c + 6d)}{(a - 3b) + (2c - 6d)} = \frac{(a + 3b) - (2c + 6d)}{(a - 3b) - (2c - 6d)}$$
2. or, $$\frac{(a + 3b) + (2c + 6d)}{(a + 3b) - (2c + 6d)} = \frac{(a - 3b) + (2c - 6d)}{(a - 3b) - (2c - 6d)}$$ [By alternendo]
3. or, $$\frac{[(a + 3b) + (2c + 6d)] + [(a + 3b) - (2c + 6d)]}{[(a + 3b) + (2c + 6d)] - [(a + 3b) - (2c + 6d)]} = \frac{[(a - 3b) + (2c - 6d)] + [(a - 3b) - (2c - 6d)]}{[(a - 3b) + (2c - 6d)] - [(a - 3b) - (2c - 6d)]}$$ [By componendo and dividendo]
4. or, $$\frac{2(a + 3b)}{2(2c + 6d)} = \frac{2(a - 3b)}{2(2c - 6d)}$$
5. or, $$\frac{a + 3b}{2(c + 3d)} = \frac{a - 3b}{2(c - 3d)}$$
6. or, $$\frac{a + 3b}{c + 3d} = \frac{a - 3b}{c - 3d}$$ [Multiplying both sides by 2]
7. or, $$\frac{a + 3b}{a - 3b} = \frac{c + 3d}{c - 3d}$$ [By alternendo]
8. or, $$\frac{(a + 3b) + (a - 3b)}{(a + 3b) - (a - 3b)} = \frac{(c + 3d) + (c - 3d)}{(c + 3d) - (c - 3d)}$$ [By componendo and dividendo]
9. or, $$\frac{2a}{6b} = \frac{2c}{6d}$$
10. or, $$\frac{a}{3b} = \frac{c}{3d}$$
11. or, $$\frac{a}{b} = \frac{c}{d}$$ [Multiplying both sides by 3]

Hence proved.