Question

In fig 6.20, de || oq and df || or. show that ef || qr.​

Answer 1

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Answer 2

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Given,

In ΔPQO, DE || OQ

So by using Basic Proportionality Theorem,

PD/DO = PE/EQ…………(i)

Again given, in ΔPQO, DE || OQ ,

So by using Basic Proportionality Theorem,

PD/DO = PF/FR………… (ii)

From equation (i) and (ii), we get,

PE/EQ = PF/FR

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ΔPQR.