If ABCD is a rhombus whose diagonals cut at the origin O, then $\vec{OA}+\vec{OB}+\vec{OC} +\vec{OD}$ equals

$\vec {AB}+\vec {AC}$ $\vec 0$ $2\vec {AB}+\vec {BC}$ $\vec {AC}+\vec {BD}$

$\vec 0$

Since the diagonals of a rhombus bisect each other. $∴\vec{OA}=-\vec{OC}$ and $\vec{OB}=-\vec{OD}$ $⇒\vec{OA}+\vec{OB}+\vec{OC} +\vec{OD}=\vec 0$