在 $△ABC$中， $\stackrel{\rightharpoonup }{Ab}=\stackrel{\rightharpoonup }{c}$， $\stackrel{\rightharpoonup }{AC}=\stackrel{\rightharpoonup }{b}$．若点 $D$满足 $\stackrel{\rightharpoonup }{BD}=2\stackrel{\rightharpoonup }{DC}$，则 $\stackrel{\rightharpoonup }{AD}=$(  )

A．

$\frac{2}{3}\stackrel{\rightharpoonup }{b}+\frac{1}{3}\stackrel{\rightharpoonup }{c}$

B．

$\frac{5}{3}\stackrel{\rightharpoonup }{b}-\frac{2}{3}\stackrel{\rightharpoonup }{c}$

C．

$\frac{2}{3}\stackrel{\rightharpoonup }{b}-\frac{1}{3}\stackrel{\rightharpoonup }{c}$

D．

$\frac{1}{3}\stackrel{\rightharpoonup }{b}+\frac{2}{3}\stackrel{\rightharpoonup }{c}$