Question

Any point X inside ΔDEF is joined to its vertices.From a point P in DX, PQ is drawn parallel to DE meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR ǁ DF. ​

Answer 1

Step-by-step explanation:

Given A△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||DE and QR||EF.

To prove PR||DF

Construction joins PR.

Proof In △XED, we have

PQ∣∣DE

∴

PD

XP

=

QE

XQ

..........(i)

In △XEF, we have

QR∣∣EF

∴

QE

XQ

=

RF

XR

..........(ii)

From (i) and (ii), we have

PD

XP

=

RF

XR

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 have

PR∣∣DE