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.
In ΔABC, D and E are the two points of AB and AC respectively,
such that, AD/DB = AE/EC.
DE || BC
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.