Question

In the given figure point P, Q and R are the points on XD, XE and XF respectively such that PQ||DE and PR||DF. Prove that QR||EF.
let's See whose gonna Ans this! ​

Answer 1

Answer:

I don't know the answer sorry

Answer 2

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∥DE and QR∥EF

To prove: PR∥DF

Proof:

Let us join PR

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 get PR∥DE.

IN THIS ANS THE VARIABLES ARE DIFFERENT. SO DO OCCURRINGLY