Question

In Fig., $DE||OQ$ and $DF||OR$. Show that $EF||QR$.

Answer 1

In $△POQ,$
$DE\parallel OQ$
$\therefore$ By basic proportionality theorem,

$\frac{PE}{EQ}=\frac{PD}{DO}............\left(1\right)$
Similarly,
In $△POR,$
$DF\parallel OR$
$\therefore$ By basic proportionality theorem,

$\frac{PD}{DO}=\frac{PF}{FR}............\left(2\right)$
$\therefore$ from $\left(1\right)$ and $\left(2\right)$,
$\frac{PE}{EQ}=\frac{PF}{FR}$
$\therefore$ By converse of Basic Proportionality Theorem,

$EF\parallel QR$