Let D, E, F be the middle points of the sides BC, CA, AB respectively of a triangle ABC. Then, $\vec{AD}+ \vec{BE} + \vec{CF}$ equals

$\vec 0$

Let the position vectors of A, B, C be $\vec a, \vec b$ and $\vec c$ respectively. Then, the position vectors of D, E and F are

$\frac{\vec b+\vec c}{2},\frac{\vec c+\vec a}{2}$ and $\frac{\vec a+\vec b}{2}$ respectively.

$\vec{AD}+ \vec{BE} + \vec{CF}$

$=\left\{\frac{\vec b+\vec c}{2}-\vec a\right\}+\left\{\frac{\vec c+\vec a}{2}-\vec b\right\}+\left\{\frac{\vec a+\vec b}{2}-\vec c\right\}=\vec 0$