In the figure, DE||OQ and DF||OR, show that EF||QR.

Given,
In \(\triangle PQO\), DE || OQ

So by using Basic Proportionality Theorem,
\(\frac{PD}{DO} = \frac{PE}{EQ}\) ………………. (i)

Again given, in \(\triangle PQO\), DE || OQ ,

So by using Basic Proportionality Theorem,
\(\frac{PD}{DO} = \frac{PF}{FR}\) ………………… (ii)

From equation (i) and (ii), we get,
\(\frac{PE}{EQ} = \frac{PF}{FR}\)

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in \(\triangle PQR\).