In figure, DE

Solution:

In figure, $D E \| O Q$ and DF $\| O R$, then by Basic Proportionality Theorem,

We have $\quad \frac{P E}{E Q}=\frac{P D}{D O}....(1)

and $\quad \frac{P F}{F R}=\frac{P D}{D O}$\ldots(2)$

From $(1)$ and $(2), \quad \frac{P E}{E Q}=\frac{P F}{F R}$

Now, in $\triangle \mathrm{PQR}$, we have proved that

$\Rightarrow \frac{P E}{E Q}=\frac{P F}{F R}$

$\mathrm{EF} \| \mathrm{QR}$

(By converse of Basic Proportionality Theorem)