In the figure,$DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.
(N/A) In $\Delta POQ$,$DE \parallel OQ$.
Therefore,$\frac{PE}{EQ} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(i)$
In $\Delta POR$,$DF \parallel OR$.
Therefore,$\frac{PF}{FR} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(ii)$
From $(i)$ and $(ii)$,we obtain:
$\frac{PE}{EQ} = \frac{PF}{FR}$
Therefore,$EF \parallel QR$ (Converse of Basic Proportionality Theorem).
Therefore,$\frac{PE}{EQ} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(i)$
In $\Delta POR$,$DF \parallel OR$.
Therefore,$\frac{PF}{FR} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(ii)$
From $(i)$ and $(ii)$,we obtain:
$\frac{PE}{EQ} = \frac{PF}{FR}$
Therefore,$EF \parallel QR$ (Converse of Basic Proportionality Theorem).
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