Question

using theorem Converse of basic proportionality theorem prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side recall that your class done it in class 9.

Answer 1

Converse of Basic Proportionality Theorem

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. 

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 

Hence the Converse of Basic Proportionality therorem is proved.

Answer 2

consider ∆ ABC in which D and E are the mid points of sides AB and AC respectively.


so,

AD / DB = 1

and AE / EC = 1

= > AD/ DB = AE/ EC

DE || BC


[ by Converse of basic proportionality theorem is the line joining the midpoints of any two side of a triangle is parallel to the third side ]