Question

prove converse of basic proportionality theorem ​

Answer 1

Answer:

If a line divides any two side of a triangle in the same ratio then the line is parallel to the third side.

in the figure line l interact the side AB and the side AC of triangle ABC in the point p and Q respectively and AP/PB = AQ/QC ,

hence line l is parallel seg BC

Answer 2

hi mate,

Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

If

AD AE

---- = ------ then DE || BC

DB EC

Given : A Δ ABC and a line intersecting AB in D and AC in E,

such that AD / DB = AE / EC.

Prove that : DE || BC

Let DE is not parallel to BC. Then there must be another line that is parallel to BC.

Let DF || BC.1) DF || BC 1) By assumption

2) AD / DB = AF / FC 2) By Basic Proportionality theorem

3) AD / DB = AE /EC 3) Given

4) AF / FC = AE / EC 4) By transitivity (from 2 and 3)

5) (AF/FC) + 1 = (AE/EC) + 1 5) Adding 1 to both side

6) (AF + FC )/FC = (AE + EC)/EC 6) By simplifying

7) AC /FC = AC / EC 7) AC = AF + FC and AC = AE + EC

8) FC = EC 8) As the numerator are same so denominators are equal

This is possible when F and E are same. So DF is the line DE itself.

∴ DF || BC

i hope it helps you.