Lets have △ABC with line DE∥BC
We have to prove that AD/DB=AE/EC
construct or draw h1 from E perpendicular to AD, and h2 from D perpendicular to AE.
Draw BE and CD.
Area △ADE/Area △BDE =1/2AD.h1/[1/2DB.h1
=AD/DB Area △ADE/Area △CED=1/2AE.h2/[1/2EC.h2]=AE/EC
but Area △BDE=Area △CED (equal base and height)
hence Area △ADE/Area △BDE = Area △ADE/Area △CED
Therefore AD/DB = AE/EC
Repeating the same method, we can show that:
AD/AB = AE/AC and AB/BD = AC/CE
Therefore since there is equal ratio on divided sides, the line is parallel to the third side of a triangle.