Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. be sure to create and name the appropriate geometric figures. this figure does not need to be submitted.

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

For a 0.25 pfor b o.3pp (0.25+0.3)=0.28