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

(i) (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd)

(ii) (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)

(iii) (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²)

(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 (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd), we get :

$\Rightarrow \dfrac{a^2 + ab}{c^2 + cd} \\[1em] \Rightarrow \dfrac{(ck)^2 + ck.dk}{c^2 + cd} \\[1em] \Rightarrow \dfrac{k^2 c^2 + k^2 cd}{c^2 + cd} \\[1em] \Rightarrow \dfrac{k^2(c^2 + cd)}{c^2 + cd} \\[1em] \Rightarrow k^2.$

Substituting value of a and b in R.H.S. of (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd), we get :

$\Rightarrow \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{(kd)^2 - 2 \times ck \times dk}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{k^2 d^2 - 2k^2 cd}{d^2 - 2cd} \\[1em] \Rightarrow \dfrac{k^2(d^2 - 2cd)}{d^2 - 2cd} \\[1em] \Rightarrow k^2.$

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

Hence, proved that (a² + ab) : (c² + cd) = (b² − 2ab) : (d² − 2cd).

(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 (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)

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

Substituting value of a and b in R.H.S. of (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc)

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

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

Hence, proved that (a² + b²) : (c² + d²) = (ab + ad − bc) : (cd − ad + bc).

(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 value of a and b in L.H.S. of (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²),

$\Rightarrow \dfrac{a^2 + ac + c^2}{a^2 - ac + c^2} \\[1em] \Rightarrow \dfrac{(kc)^2 + (kc)c + c^2}{(kc)^2 - (kc)c + c^2} \\[1em] \Rightarrow \dfrac{k^2 c^2 + kc^2 + c^2}{k^2 c^2 - kc^2 + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^2 + k + 1)}{c^2(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}.$

Substituting value of a and b in R.H.S. of (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²),

$\Rightarrow \dfrac{b^2 + bd + d^2}{b^2 - bd + d^2} \\[1em] \Rightarrow \dfrac{(kd)^2 + (kd)d + d^2}{(kd)^2 - (kd)d + d^2} \\[1em] \Rightarrow \dfrac{k^2 d^2 + kd^2 + d^2}{k^2 d^2 - kd^2 + d^2} \\[1em] \Rightarrow \dfrac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}.$

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

Hence, proved that (a² + ac + c²) : (a² − ac + c²) = (b² + bd + d²) : (b² − bd + d²).