Question

$X$ is any point inside a $△DEF$ and is joined to all vertices. From point $P$ on $DX$, $PQ||DE$ is drawn which meets $EX$ at $Q$ and from $Q,QR||EF$ is drawn which meets $XF$ at $R$. Prove that $PR||DF$

Answer 1

$\begin{array}{c}In\Delta DXE\\ PQ\parallel DE\\ so,\frac{XP}{PD}=\frac{XQ}{DE}-eq1\\ In\Delta XEF\\ QR\parallel EF\\ so,\frac{XQ}{QE}=\frac{XR}{RF}-eq2\\ NowIn\Delta DXF\\ \frac{XP}{PD}=\frac{XR}{RF}-\left(fromeq1\&2\right)\\ PR\parallel DF\end{array}$