Question

what is converse of BPT theorem?

Answer 1

hello
just a diagram is not there

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. 

Diagram
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 

   
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Hence the Converse of Basic Proportionality therorem is proved.

Answer 2

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.

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.

Statements Reasons
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