Question

If $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$$, prove that $$(b^2 + d^2 + f^2)(a^2 + c^2 + e^2) = (ab + cd + ef)^2$$.

Answer 1

Given: $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$$

To prove: $$(b^2 + d^2 + f^2)(a^2 + c^2 + e^2) = (ab + cd + ef)^2$$

Proof:

1. Let $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$$, which gives $$a = bk$$, $$c = dk$$, $$e = fk$$.
2. $$\text{L.H.S.} = (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) = (b^2 + d^2 + f^2)(b^2 k^2 + d^2 k^2 + f^2 k^2)$$
3. $$\text{L.H.S.} = k^2(b^2 + d^2 + f^2)^2$$
4. $$\text{R.H.S.} = (ab + cd + ef)^2 = ((bk)b + (dk)d + (fk)f)^2 = (b^2 k + d^2 k + f^2 k)^2$$
5. $$\text{R.H.S.} = [k(b^2 + d^2 + f^2)]^2 = k^2(b^2 + d^2 + f^2)^2$$
6. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.