In the given figure, $DE \| OQ$ and $DF \| OR$. Show that $EF \| QR$.
"
Given:
$DE \| OQ$ and $DF \| OR$.
To do:
We have to show that $EF \| QR$.
Solution:
We know that,
If a line divides two sides of a triangle proportionally, then it is parallel to the third side.
In $\triangle POQ, DE \| OQ$,
This implies,
$\frac{PE}{EQ}=\frac{PD}{DO}$.........(i)
In $\triangle POR, DF \| OR$,
This implies,
$\frac{PF}{FR}=\frac{PD}{DO}$.........(ii)
From (i) and (ii), we get,
$\frac{PE}{EQ}=\frac{PF}{FR}$
By converse of B.P.T.,
$EF \| QR$
Hence proved.
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