Question

If $$(a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d)$$, prove that $$a : b = c : d$$.

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

Answer 1

Given: $$(a + b + c + d) : (a + b - c - d) = (a - b + c - d) : (a - b - c + d)$$

To prove: $$a : b = c : d$$

Proof:

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

Hence proved.