Question

In the figure , X is any point in the interior of triangle . Point x is joined to the vertices of triangle
$segPQ\parallel segDE$
$segPR\parallel segDF$

proove:

$segQR\parallel segEF$

Answer 1

In $△DXE$

$PQ\parallel DE$

$\frac{XP}{PD}=\frac{XQ}{QE}$...(1)

In $△DXF$

$PR\parallel DF$

$\frac{XP}{PD}=\frac{XR}{RF}$...(2)

From (1) and (2)

$\frac{XQ}{QE}=\frac{XR}{RF}$

$\therefore QR\parallel EF$