If $a, b, c, d$ are in proportion, prove that :

$\begin{array}{ll} \text{(i)} & (5a + 7b)(2c - 3d) \ & = (5c + 7d)(2a - 3b) \\ \text{(ii)} & (ma + nb) : b \ & = (mc + nd) : d \\ \text{(iii)} & (a^4 + c^4) : (b^4 + d^4) \ & = a^2c^2 : b^2d^2 \\ \text{(iv)} & \dfrac{a^2 + ab}{c^2 + cd} = \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \text{(v)} & \dfrac{(a + c)^3}{(b + d)^3} = \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \text{(vi)} & \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \ & = \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2} \\ \text{(vii)} & \dfrac{a^2 + b^2}{c^2 + d^2} = \dfrac{ab + ad - bc}{bc + cd - ad} \\[1em] \text{(viii)} & abcd\Big(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2} \Big) \ & = a^2 + b^2 + c^2 + d^2. \end{array}$

(i) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = (5a + 7b)(2c - 3d) \\[0.5em] = (5bk + 7b)(2dk - 3d) \\[0.5em] = bd(5k + 7)(2k - 3) \\[1em] \text{R.H.S.} = (5c + 7d)(2a - 3b) \\[0.5em] = (5dk + 7d)(2bk - 3b) \\[0.5em] = bd(5k + 7)(2k - 3).$

Since, L.H.S. = R.H.S. Hence proved.

(ii) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = (ma + nb) : b \\[0.5em] = \dfrac{ma + nb}{b} \\[0.5em]$

Putting value of a = bk,

$= \dfrac{mbk + nb}{b} \\[0.5em] = mk + n. \\[1em] \text{R.H.S.} = (mc + nd) : d \\[0.5em]$

Putting value of c = dk,

$= \dfrac{mdk + nd}{d} \\[0.5em] = mk + n.$

Since, L.H.S. = R.H.S. Hence proved.

(iii) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = (a^4 + c^4) : (b^4 + d^4) \\[1em] = \dfrac{a^4 + c^4}{b^4 + d^4} \\[1em] = \dfrac{k^4b^4 + k^4d^4}{b^4 + d^4} \\[1em] = k^4. \\[1em] \text{R.H.S.} = a^2c^2 : b^2d^2 \\[1em] = \dfrac{a^2c^2}{b^2d^2} \\[1em] = \dfrac{(bk)^2(dk)^2}{b^2d^2} \\[1em] = \dfrac{k^4b^2d^2}{b^2d^2} \\[1em] = k^4.$

Since, L.H.S. = R.H.S. Hence proved.

(iv) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = \dfrac{a^2 + ab}{c^2 + cd} \\[1em] = \dfrac{b^2k^2 + b^2k}{d^2k^2 + d^2k} \\[1em] = \dfrac{b^2k(k + 1)}{d^2k(k + 1)} \\[1em] = \dfrac{b^2}{d^2} \\[1em] \text{R.H.S.} = \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] = \dfrac{b^2 - 2b^2k}{d^2 - 2d^2k} \\[1em] = \dfrac{b^2(1 - 2k)}{d^2(1 - 2k)} \\[1em] = \dfrac{b^2}{d^2}.$

Since, L.H.S. = R.H.S. Hence proved.

(v) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = \dfrac{(a + c)^3}{(b + d)^3} \\[1em] = \dfrac{(bk + dk)^3}{(b + d)^3} \\[1em] = \dfrac{k^3(b + d)^3}{(b + d)^3} \\[1em] = k^3 \\[1em] \text{R.H.S.} = \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] = \dfrac{bk(bk - dk)^2}{b(b - d)^2} \\[1em] = \dfrac{bk(k^2(b - d)^2)}{b(b - d)^2} \\[1em] = \dfrac{bk^3(b - d)^2}{b(b - d)^2} \\[1em] = k^3.$

Since, L.H.S. = R.H.S. Hence proved.

(vi) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] = \dfrac{b^2k^2 + (bk)b + b^2}{b^2k^2 - (bk)b + b^2} \\[1em] = \dfrac{b^2k^2 + b^2k + b^2}{b^2k^2 - b^2k + b^2} \\[1em] = \dfrac{b^2(k^2 + k + 1)}{b^2(k^2 - k + 1)} \\[1em] = \dfrac{k^2 + k + 1}{k^2 - k + 1} \\[1em] \text{R.H.S.} = \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2} \\[1em] = \dfrac{d^2k^2 + (dk)d + d^2}{d^2k^2 - (dk)d + d^2} \\[1em] = \dfrac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] = \dfrac{k^2 + k + 1}{k^2 - k + 1}.$

Since, L.H.S. = R.H.S. Hence proved.

(vii) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = \dfrac{a^2 + b^2}{c^2 + d^2} \\[1em] = \dfrac{b^2k^2 + b^2}{d^2k^2 + d^2} \\[1em] = \dfrac{b^2(k^2 + 1)}{d^2(k^2 + 1)} \\[1em] = \dfrac{b^2}{d^2} \\[1em] \text{R.H.S.} = \dfrac{ab + ad - bc}{bc + cd - ad} \\[1em] = \dfrac{(bk)b + (bk)d - b(dk)}{b(dk) + (dk)d - (bk)d} \\[1em] = \dfrac{b^2k + bdk - bdk}{bdk + d^2k - bdk} \\[1em] = \dfrac{b^2k}{d^2k} \\[1em] = \dfrac{b^2}{d^2}.$

Since, L.H.S. = R.H.S. Hence proved.

(viii) a, b, c, d are in proportion

$\therefore \dfrac{a}{b} = \dfrac{c}{d} = k$

Hence, a = bk, c = dk.

$\text{L.H.S.} = abcd\Big(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2} \Big) \\[1em] = (bk)b(dk)d\Big(\dfrac{1}{b^2k^2} + \dfrac{1}{b^2} + \dfrac{1}{d^2k^2} + \dfrac{1}{d^2}\Big) \\[1em] = b^2d^2k^2\Big(\dfrac{d^2 + d^2k^2 + b^2 + b^2k^2}{b^2d^2k^2}\Big) \\[1em] = d^2(1 + k^2) + b^2(1 + k^2) \\[1em] = (d^2 + b^2)(1 + k^2). \\[1em] \text{R.H.S.} = a^2 + b^2 + c^2 + d^2 \\[1em] = b^2k^2 + b^2 + d^2k^2 + d^2 \\[1em] = b^2(k^2 + 1) + d^2(k^2 + 1) \\[1em] = (d^2 + b^2)(1 + k^2).$

Since, L.H.S. = R.H.S. Hence proved.