Question

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

Answer 1

Given: $$(ma + nb) : (mc + nd) = (ma - nb) : (mc - nd)$$

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

Proof:

1. $$\frac{ma + nb}{mc + nd} = \frac{ma - nb}{mc - nd}$$ [Given]
2. $$\frac{ma + nb}{ma - nb} = \frac{mc + nd}{mc - nd}$$ [By alternendo]
3. $$\frac{(ma + nb) + (ma - nb)}{(ma + nb) - (ma - nb)} = \frac{(mc + nd) + (mc - nd)}{(mc + nd) - (mc - nd)}$$ [By componendo and dividendo]
4. $$\frac{2ma}{2nb} = \frac{2mc}{2nd}$$
5. $$\frac{ma}{nb} = \frac{mc}{nd}$$
6. $$\frac{a}{b} = \frac{c}{d}$$ [Multiplying both sides by $$\frac{n}{m}$$]
7. $$a : b = c : d$$

Hence proved.