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# $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \| QR$:

**(i)** $PE = 3.9\ cm, EQ = 3\ cm, PF = 3.6\ cm$ and $FR = 2.4\ cm$

**(ii)** $PE = 4\ cm, QE = 4.5\ cm, PF = 8\ cm$ and $RF = 9\ cm$

**(iii)** $PQ = 1.28\ cm, PR = 2.56\ cm, PE = 0.18\ cm$ and $PF = 0.36\ cm$

To do:

We have to find whether $EF \parallel QR$ in each case.

Solution:

(i) We know that,

If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

$\frac{PE}{EQ}=\frac{3.9}{3}=1.3$

$\frac{PF}{FR}=\frac{3.6}{2.4}=\frac{3}{2}$

$\frac{PE}{EQ}≠\frac{PF}{FR}$

Hence, by converse of proportionality theorem $EF$ is not parallel to $QR$.

(ii) We know that,

If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}$

$\frac{PF}{FR}=\frac{8}{9}$

$\frac{PE}{EQ}=\frac{PF}{FR}$

Hence, by converse of proportionality theorem $EF$ is parallel to $QR$.

(iii) We know that,

If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

$\frac{PQ}{PE}=\frac{1.28}{0.18}=\frac{64}{9}$

$\frac{PR}{PF}=\frac{2.56}{0.36}=\frac{64}{9}$

$\frac{PQ}{PE}=\frac{PR}{PF}$

Hence, by converse of proportionality theorem $EF$ is parallel to $QR$.

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