设 $D$为 $△ABC$所在平面内一点 $\stackrel{\rightharpoonup }{BC}=3\stackrel{\rightharpoonup }{CD}$，则（  ）

A．	$\stackrel{\rightharpoonup }{AD}=\frac{1}{3}\stackrel{\rightharpoonup }{AB}+\frac{4}{3}\stackrel{\rightharpoonup }{AC}$	B．	$\stackrel{\rightharpoonup }{AD}=\frac{1}{3}\stackrel{\rightharpoonup }{AB}-\frac{4}{3}\stackrel{\rightharpoonup }{AC}$
C．	$\stackrel{\rightharpoonup }{AD}=\frac{4}{3}\stackrel{\rightharpoonup }{AB}+\frac{1}{3}\stackrel{\rightharpoonup }{AC}$	D．	$\stackrel{\rightharpoonup }{AD}=\frac{4}{3}\stackrel{\rightharpoonup }{AB}-\frac{1}{3}\stackrel{\rightharpoonup }{AC}$