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

(i) Given, in Triangle ABC, DE II BC
\({{AD} \over {DB}} = {{AE} \over {EC}}\)
[Using Basic proportionality theorem]
\(\Rightarrow {{1.5} \over {3}} = {{1} \over {EC}}\)
\(\Rightarrow EC = {{3} \over {1.5}}\)
\(\Rightarrow \)EC = 3×\({{10} \over {15}}\) = 2 cm
Hence, EC = 2 cm.


(ii) Given, in Triangle ABC, DE is parallel to BC
= \({{AD} \over {DB}} = {{AE} \over {EC}}\)
[Using Basic proportionality theorem]
\(\Rightarrow {{AD} \over {7.2}}= {{1.8} \over {5.4}}\)
\(\Rightarrow AD = 1.8 ×{{7.2} \over {5.4}} \)
\( = ({{18} \over {10}})×({{72} \over {10}})×({{10} \over {54}}) \)
\( = {{24} \over {10}}\)

AD = 2.4 cm

Hence, AD = 2.4 cm.