Question

prove that if a line divides any two sodes of a triangle in same ratio, the line parallel to third side. ​

Answer 1

Answer:

According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. Assume DE is not parallel to BC. Now, draw a line DE' parallel to BC. This is possible only when E and E' coincides.

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Answer 2

Step-by-step explanation:

Given : The line l intersects the sides PQ and side PR of ΔPQR in the points M and Nrespectively such that MQPM=NRPN and P−M−Q, P−N−R.

To Prove : Line l ∥ Side QR

Proof : Let us consider that line l is not parallel to the side QR. Then there must be another line passing through M which is parallel to the side QR.

Let line MK be that line.

Line MK intersects the side PR at K, (P−K−R)

In ΔPQR, line MK∥ side QR

∴   MQPM=KRPK          ....(1) (B.P.T.)

But MQPM=NRPN             ....(2) (Given)

∴ KRPK=NRPN    [From (1) and (2)]

∴ KRPK+KR=NRPN+NR     (P−K−R and P−N−R)

∴ the points K and N are not different.

∴ line MK and line MN coincide

∴ line MN∥ Side