In the adjoining figure, DE ∥ BC.

(i) If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x.

(ii) If DB = x - 3, AB = 2x, EC = x - 2 and AC = 2x + 3, find the value of x.

(i) Considering △ABC and △ADE,

∠A = ∠A (Common angle)
∠ADE = ∠ABC (Corresponding angles)

So, by AA rule of similarity △ABC ~ △ADE. Since ratio of corresponding sides is same,

$\therefore \dfrac{AD}{AB} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \\[1em] \Rightarrow \dfrac{x}{x + x - 2} = \dfrac{x + 2}{x + 2 + x - 1} \\[1em] \Rightarrow \dfrac{x}{2x - 2} = \dfrac{x + 2}{2x + 1} \\[1em] \Rightarrow x(2x + 1) = (x + 2)(2x - 2) \\[1em] \Rightarrow 2x^2 + x = 2x^2 - 2x + 4x - 4 \\[1em] \Rightarrow 2x^2 - 2x^2 + x + 2x - 4x = 4 \\[1em] \Rightarrow -x = -4 \\[1em] \Rightarrow x = 4.$

Hence, the value of x = 4.

(ii) Since, triangles are similar so, ratio of corresponding sides is same,

$\therefore \dfrac{AD}{AB} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{AB - DB}{AB} = \dfrac{AC - EC}{AC} \\[1em] \Rightarrow \dfrac{2x - (x - 3)}{2x} = \dfrac{2x + 3 - (x - 2)}{2x + 3} \\[1em] \Rightarrow \dfrac{x + 3}{2x} = \dfrac{x + 5}{2x + 3} \\[1em] \Rightarrow (x + 3)(2x + 3) = 2x(x + 5) \\[1em] \Rightarrow 2x^2 + 3x + 6x + 9 = 2x^2 + 10x \\[1em] \Rightarrow 2x^2 - 2x^2 + 9 = 10x - 9x \\[1em] \Rightarrow x = 9 \\[1em]$

Hence, the value of x = 9.