In figure,

Solution:

(i) In figure, (i) DE $\| \mathrm{BC}$ (Given)

$\Rightarrow \frac{A D}{D B}=\frac{A E}{E C}$ (By Basic Proportionality Theorem)

$\Rightarrow \frac{1.5}{3}=\frac{1}{E C}$

$\{\because \mathrm{AD}=1.5 \mathrm{~cm}, \mathrm{DB}=3 \mathrm{~cm}$ and $\mathrm{AE}=1 \mathrm{~cm}\}$

$\Rightarrow \mathrm{EC}=\frac{3}{1.5}=2 \mathrm{~cm}$

(ii) In fig. (ii) $D E \| B C$ (given)

So, $\frac{A D}{B D}=\frac{A E}{C E} \Rightarrow \frac{A D}{7.2}=\frac{1.8}{5.4}$

$\{\because \mathrm{BD}=7.2, \mathrm{AE}=1.8 \mathrm{~cm}$ and $\mathrm{CE}=5.4 \mathrm{~cm}\}$

$\mathrm{AD}=2.4 \mathrm{~cm}$