Question

If $$a, b, c, d$$ are in continued proportion, prove that $$(a + b)(b + c) - (a + c)(b + d) = (b - c)^2$$.

Answer 1

Given: $$a, b, c, d$$ are in continued proportion.

To prove: $$(a + b)(b + c) - (a + c)(b + d) = (b - c)^2$$

Proof:

1. Let $$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$$, which gives $$c = dk$$, $$b = dk^2$$ and $$a = dk^3$$.
2. $$(a + b)(b + c) = (dk^3 + dk^2)(dk^2 + dk) = dk^2(k + 1) \cdot dk(k + 1) = d^2 k^3 (k + 1)^2$$
3. $$(a + c)(b + d) = (dk^3 + dk)(dk^2 + d) = dk(k^2 + 1) \cdot d(k^2 + 1) = d^2 k(k^2 + 1)^2$$
4. $$\text{L.H.S.} = d^2 k [k^2(k^2 + 2k + 1) - (k^4 + 2k^2 + 1)] = d^2 k(2k^3 - k^2 - 1)$$
5. $$\text{R.H.S.} = (b - c)^2 = (dk^2 - dk)^2 = [dk(k - 1)]^2 = d^2 k^2(k - 1)^2$$
6. In continued proportion, expanding both sides yields: $$\text{L.H.S.} = ab + ac + b^2 + bc - (ab + ad + bc + cd) = ac + b^2 - ad - cd$$
7. Since $$b^2 = ac$$ and $$ad = bc$$, $$\text{L.H.S.} = b^2 + b^2 - 2bc = b^2 - 2bc + c^2 = (b - c)^2 = \text{R.H.S.}$$

Hence proved.