maths class10 chapter 6 theorems converse of basic proportionality theory
Answers
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.e.
∴ DE || BC
i hope it helps you.