Question

convose of the basic proportionality thheorm

Answer 1

Converse of Basic Proportionality Theorem
Statement
If a line divides any two sides of a triangle (Δ) in the same ratio, then the line must be parallel (||) to the third side.
Given
In ΔABC, D and E are the two points of AB and AC respectively,
such that, AD/DB = AE/EC.

To Prove
DE || BC

Proof
In ΔABC,
given, AD/DB = AE/EC ----- (1)

Let us assume that in ΔABC, the point F is an intersect on the side AC.
So we can apply the Thales Theorem, AD/DB = AF/FC ----- (2)

Simplify, in (1) and (2) ==> AE/EC = AF/FC
Add 1 on both sides, ==> (AE/EC) + 1 = (AF/FC) + 1
==> (AE+EC)/EC = (AF+FC)/FC
==> AC/EC = AC/FC
==> EC = FC
From the above, we can say that the points E and F coincide on AC.
i.e., DF coincides with DE.

Since DF is parallel to BC, DE is also parallel BC


Hence the Converse of Basic Proportionality therorem is proved.

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.e.

∴ DE || BC

i hope it helps you.