If a, b, c and d are in proportion, prove that:
(i) (5a + 7v) (2c – 3d) = (5c + 7d) (2a – 3b)
(ii) (ma + nb): b = (mc + nd): d
(iii) \left(a^{4}+c^{4}\right):\left(b^{4}+d^{4}\right)=a^{2} c^{2}: b^{2} d^{2}
(iv) \frac{a^{2}+a b}{c^{2}+c d}=\frac{b^{2}-2 a b}{d^{2}-2 c d}
(v) \frac{(\mathbf{a}+\mathbf{c})^{3}}{(\mathbf{b}+\mathbf{d})^{3}}=\frac{\mathbf{a}(\mathbf{a}-\mathbf{c})^{2}}{\mathbf{b}(\mathbf{b}-\mathbf{d})^{2}}
(vi) \frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}=\frac{c^{2}+c d+d^{2}}{c^{2}-c d+d^{2}}
(vii) \frac{a^{2}+b^{2}}{c^{2}+d^{2}}=\frac{a b+a d-b c}{b c+c d-a d}
(viii) \operatorname{abcd}\left[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right]=a^{2}+b^{2}+c^{2}+d^{2}