Question

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

Answer 1

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

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

Proof:

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

Hence proved.