If $D, E, F$ are the midpoints of the sides $B C, C A, A B,$ respectively, of a triangle $A B C$ and $O$ is any point, then

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Solution

Let the position vector of $A, B, C, D, E, F$ be $a, b, c, d, e, f$ . So
$d = \frac{b + c}{2} c = \frac{c + a}{2} f = \frac{a + b}{2}$
$O D + O E + O F = d + e + f = \frac{b + c}{2} + \frac{c + a}{2} + \frac{a + b}{2}$
$= a + b + c$
A) $A D + B F + C F = (d - a) + (e - b) + (f - c) = (d + e + f) - (a + b + c) = 0$
B) $O A + O B + O C = a + b + c = d + e + f = O D + O E + O F$
C) $O E + O F + D O = e + f - d = a = O A$
D) $A D + \frac{2}{3} B E + \frac{1}{3} C F = (d - a) + \frac{2}{3} (e - b) + \frac{1}{3} (f - c)$
Substituting the values of $d, e, f$ and simplifying we obtain it equal to $\frac{1}{2} (c - a) = \frac{1}{2} A C$

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