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# 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]
$$\Rightarrow {{AD} \over {7.2}}= {{1.8} \over {5.4}}$$
$$\Rightarrow AD = 1.8 ×{{7.2} \over {5.4}}$$
$$= ({{18} \over {10}})×({{72} \over {10}})×({{10} \over {54}})$$
$$= {{24} \over {10}}$$