Question

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

Answer 1

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

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

Proof:

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

Hence proved.