ABC is a triangle and D,E,F are the middle points of the sides. Forces represented by AD,2\3BE and 1\3CF act on a particle at the point where AD and BE meet. Show that the resultant is represented in magnitude and direction by 1\2AC and that it's line of action divides BC in the ratio 2:1.

Given in ΔABC, D, E and F are midpoints of sides AB, BC and CA respectively.
BC = EC
Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.
Hence DF = (1/2) BC
⇒ (DF/BC) = (1/2)  → (1)
Similarly, (DE/AC) = (1/2)  → (2)
(EF/AB) = (1/2)  → (3)
From (1), (2) and (3) we have
DF/BC = DE/AC = EF/AB = 1/2 
But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar
Hence ΔABC ~ ΔEDF [By SSS similarity theorem]
(DF/BC)^2 = (1/2)^2 = 1/4 
Hence area of ΔDEF : area of ΔABC = 1 : 4