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