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

Asked by Pragya Singh | 1 year ago |  38

##### Solution :-

Given,

In ΔPQO, DE || OQ

So by using Basic Proportionality Theorem,

$$\dfrac{ PD}{DO}=\dfrac{ PE}{EQ}$$……………… ..(i)

Again given, in ΔPOR, DF || OR,

So by using Basic Proportionality Theorem,

$$\dfrac{ PD}{DO}=\dfrac{ PF}{FR}$$………………… (ii)

From equation (i) and (ii), we get,

$$\dfrac{ PE}{EQ}$$ = $$\dfrac{ PF}{FR}$$

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ΔPQR.

Answered by Abhisek | 1 year ago

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