Xis any point inside a triangle …

GIVEN A {tex} \Delta{/tex}DEF and a point X inside it. Point X is joined to the vertices D, E and F. P is any point on DX. {tex} P Q \| D E{/tex} and {tex} Q R \| E F.{/tex}
TO PROVE {tex}P R \| D F{/tex}

CONSTRUCTION Join PR.
PROOF In {tex} \Delta{/tex} XED, we have
{tex} P Q \| D E{/tex}
Therefore,by basic proportionality theorm,we have,
{tex} \therefore \quad \frac { X P } { P D } = \frac { X Q } { Q E }{/tex} ... (i)
In {tex} \Delta{/tex} XEF, we have,
{tex}Q R \| E F{/tex}
Therefore,by basic proportionality theorm,we have,
{tex}\therefore \quad \frac { \mathrm { XQ } } { \mathrm { Q } E } = \frac { \mathrm { X } R } { R F }{/tex} ... (ii) 
From (i) and (ii), we have,
{tex}\frac { X P } { P D } = \frac { X R } { R F }{/tex}
Thus, in {tex}\Delta{/tex} XFD, points R and P are dividing sides XF and XD in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have
{tex} P R \| D F{/tex}