Question

In the following figure, DE || OQ and DF || OR, show that EF || QR.

Answer 1

In ΔPOQ, DE || OQ

∴ `("PE")/("EQ") = ("PD")/("DO")`        ....(Basic Proportionality theorem)  ...(i)

In ΔPOR, DF || OR

∴ `("PF")/("FR") = ("PD")/("DO")`         ...(Basic Proportionality theorem)     ...(ii)

From (i) and (ii) we obtain

⇒ `("PE")/("EQ") = ("PD")/("DO") = ("PF")/("FR")`

⇒ `("PE")/("EQ") = ("PF")/("FR")`

∴ EF || QR                         ...(Converse of basic proportionally theorem)