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

(i) PE = 3.9 cm, EQ = 3 cm, PF = 8cm and RF = 9 cm.

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

(i) So, according to question two triangles will be formed PQR and PEF.

Comparing ratios of corresponding side we get,

$\Rightarrow \dfrac{PE}{PQ} \text{ and } \dfrac{PF}{PR} \\[1em] \Rightarrow \dfrac{PE}{PE + EQ} \text{ and } \dfrac{PF}{PF + FR} \\[1em] \Rightarrow \dfrac{3.9}{3.9 + 3} \text{ and } \dfrac{8}{8 + 9} \\[1em] \Rightarrow \dfrac{3.9}{6.9} \text{ and } \dfrac{8}{17} \\[1em] \Rightarrow \dfrac{39}{69} \text{ and } \dfrac{8}{17}.$

Since, both ratios are different hence, EF and QR are not parallel.

(ii) The triangles are shown in the figure below:

Comparing ratios of corresponding side we get,

$\Rightarrow \dfrac{PE}{PQ} \text{ and } \dfrac{PF}{PR} \\[1em] \Rightarrow \dfrac{0.18}{1.28} \text{ and } \dfrac{0.36}{2.56} \\[1em] \Rightarrow \dfrac{18}{128} \text{ and } \dfrac{36}{256} \\[1em] \Rightarrow \dfrac{9}{64} \text{ and } \dfrac{9}{64}.$

Since, both ratios are same hence by converse of basic proportionality theorem triangles are similar and so EF and QR are parallel.