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

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#### Solution

In Δ POQ, 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) = (PF)/(FR)`

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

Concept: Similarity of Triangles

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