Question

in figure 6.20 De parallel to OQ and DF parallel toOR show that EF parallel to QR​

Answer 1

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solution=>

∆PQR s ED || QO

.°. PD/DO = PE/EQ ( from formula )-------(1)

∆POR s DF || OR

.°. PD/DO = PF/FR ( from formula )........(2)

from (1) and (2) we get

PE/EQ = PF/FR

.°. EF || QR

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Answer 2

Answer:

In triangle PQA, DE parallel to AQ.

.°. PE / EQ = PD / DA ...(1) (by basic property theorem)

In triangle PRA, DF parallel to AR.

.°. PD / DA = PF / FR ...(2) (by basic property theorem)

.From (1) and (2), we get

PE / EQ = PF / FR

Thus, in triangle PQR, PE / EQ = PF / FR

EF parallel to QR