Question

The point 'E" divides the segment $PQ$ internally in the ratio $1:2$ and $R$ is any point not on the line $PQ.$ If $F$ is a point on $QR$ such that $QF:FR=2:1$ then show that $EF$ is parallel to $PR$

Answer 1

REF.Image

Let ratio factor

be x.

To prove . $PR||EF$

Now, $\frac{QE}{QP}=\frac{QF}{QR}$ and $\angle Q$ is common

$\Rightarrow \frac{EF}{PR}=\frac{QF}{QR}=\frac{QE}{QP}$

[As $△QFE\sim △QRP$ by

[SAS] congruence]

Since $\angle FQE=\angle RQP$

$EF\left|\right|PR$

As they both substends equal 4.