Converse of Basic Proportionality Theorem

If a line intersects any two sides of a triangle in equal ratio, then the line is parallel to the third side.

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.

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