Question

If $$(4a^2 + 7b^2) : (4a^2 - 7b^2) = (4c^2 + 7d^2) : (4c^2 - 7d^2)$$, prove that $$a : b = c : d$$.

Answer 1

Given: $$(4a^2 + 7b^2) : (4a^2 - 7b^2) = (4c^2 + 7d^2) : (4c^2 - 7d^2)$$

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

Proof:

1. $$\frac{4a^2 + 7b^2}{4a^2 - 7b^2} = \frac{4c^2 + 7d^2}{4c^2 - 7d^2}$$ [Given]
2. $$\frac{(4a^2 + 7b^2) + (4a^2 - 7b^2)}{(4a^2 + 7b^2) - (4a^2 - 7b^2)} = \frac{(4c^2 + 7d^2) + (4c^2 - 7d^2)}{(4c^2 + 7d^2) - (4c^2 - 7d^2)}$$ [By componendo and dividendo]
3. $$\frac{8a^2}{14b^2} = \frac{8c^2}{14d^2}$$
4. $$\frac{a^2}{b^2} = \frac{c^2}{d^2}$$ [Multiplying both sides by $$\frac{14}{8}$$]
5. $$\frac{a}{b} = \frac{c}{d}$$ [Taking square root on both sides]
6. $$a : b = c : d$$

Hence proved.