Question

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

Answer 1

It can be shown using BPT theorem

Answer 2

Answer:

HELLO MATE ,

EF//QR [CONVERSE OF BPT]

Step-by-step explanation:

GIVEN;

DE//OQ

DF//OR

PROVE;

EF//QR

SOLUTION;

IN ΔPOQ ,  DE//OQ

       $\frac{PE}{EQ} =\frac{PD}{DO} ------[BPT]--1$

IN ΔPOR ,  DF//OR

      $\frac{PE}{FR} =\frac{PF}{FR} ------[BPT]--2$

From equation 1 and 2

      $\frac{PE}{EQ} =\frac{PF}{FR}$

IN ΔPQR  ,  EF//QR

      $\frac{FE}{EQ} =\frac{PF}{FR} ------[proved]$

THEREFORE, EF//QR [converse of BPT]

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