In figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i) We know that,

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

$\therefore \dfrac{AD}{BD} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{1.5}{3}= \dfrac{1}{EC} \\[1em] \Rightarrow EC = \dfrac{3 \times 1}{1.5} \\[1em] \Rightarrow EC = 2 \text{ cm}.$

Hence, EC = 2 cm.

(ii) We know that,

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

$\therefore \dfrac{AD}{BD} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{AD}{7.2}= \dfrac{1.8}{5.4} \\[1em] \Rightarrow \dfrac{AD}{7.2} = \dfrac{1}{3} \\[1em] \Rightarrow AD = 7.2 \times \dfrac{1}{3} = 2.4 \text{ cm}.$

Hence, AD = 2.4 cm.