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# In the given figure, $DE \| BC$. Find $EC$ in (i) and $AD$ in (ii).

**(i)**

**(ii)**

"

Given:

$DE \| BC$

To do:

We have to find $EC$ in (i) and $AD$ in (ii).

Solution:

(i) In $\triangle ABC$, by B.P.T.,

$\frac{AD}{DB}=\frac{AE}{EC}$

$\frac{1.5}{3}=\frac{1}{EC}$

$EC=\frac{3}{1.5}$

$EC=2\ cm$

Therefore, $EC=2\ cm$.

(ii) In $\triangle ABC$, by B.P.T.,

$\frac{AD}{DB}=\frac{AE}{EC}$

$\frac{AD}{7.2}=\frac{1.8}{5.4}$

$AD=\frac{1\times7.2}{3}$

$AD=2.4\ cm$

Therefore, $AD=2.4\ cm$.

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