Question

Any point X inside $△DEF$ is joined to its vertices. From a point $P$ on $DX,PQ$ is drawn parallel to $DE$ meeting $XE$ at $Q$ and $QR$ is drawn parallel to $EF$ meeting $XF$ at $R.$ Prove that $PR\parallel DF.$

Answer 1

Given $△DEF$ and a point $X$ inside it. Point $X$ is joined to the vertices $D,E$ and $F.P$ is any point on $DX.PQ\parallel DE$ and $QR\parallel EF$

To prove: $PR\parallel DF$

Proof:

Let us join $PR$

In $△XED,$

we have, $PQ\parallel DE$

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

In $△XEF,$

we have, $QR\parallel EF$

$\therefore$ $\frac{XQ}{QE}=\frac{XR}{RF}$ . . . (ii)

From (i) and (ii),

we have, $\frac{XP}{PD}=\frac{XR}{RF}$

Thus, in $△XFD$, points $R$ and $P$ are dividing sides $EF$ and $XD$ in the same ratio.

Therefore, by the converse of Basic Proportionality Theorem, we get $PR\parallel DE.$