Let `D ,Ea n dF` be the middle points of the sides `B C ,C Aa n dA B ,` respectively of a triangle `A B Cdot` Then prove that ` vec A D+ vec B E+ vec C F= vec0` .

Explanation

Let the position vectors of A, B and C be `veca, vecb and vecc` respectively. Then the positive vectors of D, E and F are `(vecb + vecc)//2, (vecc + veca)//2a and (veca + vecb)//2`, respectively. Therefore,
`" "vec(AD) + vec(BE) + vec(CF) = ((vecb + vecc)/(2) - veca) + ((vecc + veca )/(2) - vecb) + ((veca + vecb)/(2) - vecc) = vec0`