Question

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

Answer 1

Given: $$a : b :: c : d$$

To prove: $$\frac{a^2 + ac + c^2}{a^2 - ac + c^2} = \frac{b^2 + bd + d^2}{b^2 - bd + d^2}$$

Proof:

1. Let $$\frac{a}{b} = \frac{c}{d} = k$$, then $$a = bk$$ and $$c = dk$$
2. $$\text{L.H.S.} = \frac{a^2 + ac + c^2}{a^2 - ac + c^2} = \frac{b^2 k^2 + (bk)(dk) + d^2 k^2}{b^2 k^2 - (bk)(dk) + d^2 k^2}$$
3. $$\text{L.H.S.} = \frac{k^2 (b^2 + bd + d^2)}{k^2 (b^2 - bd + d^2)} = \frac{b^2 + bd + d^2}{b^2 - bd + d^2}$$
4. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.