In the figure (i) given below, if DE ∥ BC, AD = 3 cm, BD = 4 cm and BC = 5 cm, find (i) AE : EC (ii) DE.

(i) Considering △ABC and △ADE,

∠ A = ∠ A (Common angle)
∠ ADE = ∠ABC (Alternate angle)

So, by AA rule of similarity △ABC ~ △ADE.

$\therefore \dfrac{AE}{AC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{AE}{AE + EC} = \dfrac{AD}{AD + BD} \\[1em] \Rightarrow \dfrac{AE}{AE + EC} = \dfrac{3}{3 + 4} \\[1em] \Rightarrow 7AE = 3(AE + EC) \\[1em] \Rightarrow 7AE = 3AE + 3EC \\[1em] \Rightarrow 7AE - 3AE = 3EC \\[1em] \Rightarrow 4AE = 3EC \\[1em] \Rightarrow \dfrac{AE}{EC} = \dfrac{3}{4}.$

Hence, AE : EC = 3 : 4.

(ii) Since, △ABC ~ △ADE

$\therefore \dfrac{DE}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{AD}{AD + BD} \\[1em] \Rightarrow \dfrac{DE}{5} = \dfrac{3}{3 + 4} \\[1em] \Rightarrow DE = \dfrac{15}{7} \\[1em] \Rightarrow DE = 2\dfrac{1}{7}. \\[1em]$

Hence, DE = $2\dfrac{1}{7}.$