#### Question

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

#### 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)

Is there an error in this question or solution?

Solution In the following figure, DE || OQ and DF || OR, show that EF || QR Concept: Similarity of Triangles.