已知 $O$是 $△ABC$所在平面内一点， $D$为 $BC$边中点，且 $2\stackrel{\rightharpoonup }{OA}+\stackrel{\rightharpoonup }{OB}+\stackrel{\rightharpoonup }{OC}=\mathbf{0}$，那么（）

A．	$\stackrel{\rightharpoonup }{AO}=\stackrel{\rightharpoonup }{OD}$	B．	$\stackrel{\rightharpoonup }{AO}=2\stackrel{\rightharpoonup }{OD}$
C．	$\stackrel{\rightharpoonup }{AO}=3\stackrel{\rightharpoonup }{OD}$	D．	$2\stackrel{\rightharpoonup }{AO}=\stackrel{\rightharpoonup }{OD}$