In Fig. 6.20, DE || OQ and DF || OR. Show thatEF || QR.

Given that In triangle POQ, DE parallel to OQ
Hence,
(Basic proportionality theorem) (i)
Now,
![]()
Hence,
(Basic proportionality theorem) (ii)
From (i) and (ii), we get
![]()
Therefore,
EF is parallel to QR (Converse of basic proportionality theorem)