**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.