Prove converse basic propotionality 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

Ex14.1