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.
