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

Q5  In Fig. 6.20, DE || OQ and DF || OR. Show that
       EF || QR. 

      

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Given : DE || OQ  and DF || OR.

To prove EF || QR. 

 Since DE || OQ so we have 

\frac{PE}{EQ}=\frac{PD}{DO}.............................................1

Also, DF || OR

\frac{PF}{FR}=\frac{PD}{DO}.............................................2

 

From equation 1 and 2, we have 

\frac{PE}{EQ} = \frac{PF}{FR }

Thus,  EF || QR.    (converse of basic proportionality theorem)

Hence proved 

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