Question

In the following figure: DE || AQ and DF || AR. Prove that EF || QR.

Answer 1

Given:

In △PQR, D lies on PA, E lies on PQ, F lies on PR, with DE || AQ and DF || AR.

Since DE || AQ, in triangles △PDE and △PQA,

∠PED = ∠PQA, ∠PDE = ∠PAQ

Therefore, △PDE ∼ △PQA

So, corresponding sides are proportional:

`(PE)/(PQ) = (PD)/(PA)`    .....(1)

Again, since DF || AR, in triangles △PDF and △PAR,

∠PFD = ∠PRA, ∠PDF = ∠PAR

Therefore, △PDF ∼ △PAR

So,

`(PF)/(PR) = (PD)/(PA)`    .....(2)

From (1) and (2),

`(PE)/(PQ) = (PF)/(PR)`

Now, in △PQR, points E and F lie on PQ and PR, respectively, and

`(PE)/(PQ) = (PF)/(PR)`

Hence, by the converse of the Basic Proportionality Theorem,

EF || QR