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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 $EF \| QR$:

$PQ = 1.28\ cm, PR = 2.56\ cm, PE = 0.18\ cm$ and $PF = 0.36\ cm$

Given:

$PQ = 1.28\ cm, PR = 2.56\ cm, PE = 0.18\ cm$ and $PF = 0.36\ cm$

To do:

We have to find if $EF\parallel QR$.

Solution:

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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