If a : b :: c : d, show that :

(i) $\dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$

(ii) $\dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}}$

(iii) $\dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}}$

(iv) $\dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}}$

(i) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)}$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of $\dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$, we get :

$\Rightarrow \dfrac{a + b}{c + d} \\[1em] \Rightarrow \dfrac{kc + kd}{c + d} \\[1em] \Rightarrow \dfrac{k(c + d)}{(c + d)} \\[1em] \Rightarrow k.$

Substituting value of a and b in R.H.S. of $\dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$, we get :

$\Rightarrow \sqrt{\dfrac{2a^2 + 7b^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\dfrac{2(kc)^2 + 7(dk)^2}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{\dfrac{k^2(2c^2 + 7d^2)}{2c^2 + 7d^2}} \\[1em] \Rightarrow \sqrt{k^2} \\[1em] \Rightarrow k.$

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

Hence, proved that $\dfrac{a + b}{c + d} = \sqrt{\dfrac{2a^{2} + 7b^{2}}{2c^{2} + 7d^{2}}}$.

(ii) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)}$

⇒ a = ck and b = dk.

Substituting value of a and b in L.H.S. of $\dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}}$, we get :

$\Rightarrow \dfrac{ma^2 + nc^2}{mb^2 + nd^2} \\[1em] \Rightarrow \dfrac{m(kc)^2 + nc^2}{m(kd)^2 + nd^2} \\[1em] \Rightarrow \dfrac{m k^2 c^2 + n c^2}{m k^2 d^2 + n d^2} \\[1em] \Rightarrow \dfrac{c^2(m k^2 + n)}{d^2(m k^2 + n)} \\[1em] \Rightarrow \dfrac{c^2}{d^2}.$

Substituting value of a and b in R.H.S. of $\dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}}$, we get :

$\Rightarrow \sqrt{\dfrac{a^4 + c^4}{b^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{(kc)^4 + c^4}{(kd)^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{k^4 c^4 + c^4}{k^4 d^4 + d^4}} \\[1em] \Rightarrow \sqrt{\dfrac{c^4(k^4 + 1)}{d^4(k^4 + 1)}} \\[1em] \Rightarrow \sqrt{\dfrac{c^4}{d^4}} \\[1em] \Rightarrow \dfrac{c^2}{d^2}.$

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

Hence, proved that $\dfrac{ma^{2} + nc^{2}}{mb^{2} + nd^{2}} = \sqrt{\dfrac{a^{4} + c^{4}}{b^{4} + d^{4}}}$.

(iii) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)}$

⇒ a = ck and b = dk.

Substituting values of a and b in L.H.S. of $\dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}}$, we get:

$\Rightarrow \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kc)(kd) + (kd)^2}{(kc)^2 - (kc)(kd) + (kd)^2} \\[1em] \Rightarrow \dfrac{k^2 c^2 + k^2 cd + k^2 d^2}{k^2 c^2 - k^2 cd + k^2 d^2} \\[1em] \Rightarrow \dfrac{k^2(c^2 + cd + d^2)}{k^2(c^2 - cd + d^2)} \\[1em] \Rightarrow \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2}.$

Substituting values of a and b in R.H.S. of $\dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}}$ ,we get:

$\Rightarrow \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2}.$

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

Hence, proved that $\dfrac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \dfrac{c^{2} + cd + d^{2}}{c^{2} - cd + d^{2}}$.

(iv) Given,

⇒ a : b :: c : d

∴ a : b = c : d

$\therefore \dfrac{a}{b} = \dfrac{c}{d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} = k \text{(let)}$

⇒ a = ck and b = dk.

Substituting values of a and b in L.H.S. of $\dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}}$, we get:

$\Rightarrow \dfrac{(a + c)^3}{(b + d)^3} \\[1em] \Rightarrow \dfrac{(kc + c)^3}{(kd + d)^3} \\[1em] \Rightarrow \dfrac{c^3(k + 1)^3}{d^3(k + 1)^3} \\[1em] \Rightarrow \dfrac{c^3}{d^3}.$

Substituting values of a and b in R.H.S. of $\dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}}$, we get:

$\Rightarrow \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \Rightarrow \dfrac{kc(kc - c)^2}{kd(kd - d)^2} \\[1em] \Rightarrow \dfrac{kc \cdot c^2(k - 1)^2}{kd \cdot d^2(k - 1)^2} \\[1em] \Rightarrow \dfrac{kc^3}{kd^3} \\[1em] \Rightarrow \dfrac{c^3}{d^3}.$

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

Hence, proved that $\dfrac{(a + c)^{3}}{(b + d)^{3}} = \dfrac{a(a - c)^{2}}{b(b - d)^{2}}$.