Cho tam giác $ABC$ và một điểm $M$ tùy ý. Mệnh đề nào sau đây đúng ?

$\overrightarrow {AI}  = \dfrac{1}{4}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).$

$\overrightarrow {AI}  = \dfrac{1}{4}\left( {\overrightarrow {AB}  - \overrightarrow {AC} } \right).$

$\overrightarrow {AI}  = \dfrac{1}{4}\overrightarrow {AB}  + \dfrac{1}{2}\overrightarrow {AC} .$

$\overrightarrow {AI}  = \dfrac{1}{4}\overrightarrow {AB}  - \dfrac{1}{2}\overrightarrow {AC} .$

$\overrightarrow {AG}  = \dfrac{2}{3}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).$

$\overrightarrow {AG}  = \dfrac{1}{3}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).$

$\overrightarrow {AG}  = \dfrac{1}{3}\overrightarrow {AB}  + \dfrac{2}{2}\overrightarrow {AC} .$

$\overrightarrow {AI}  = \dfrac{2}{3}\overrightarrow {AB}  + 3\overrightarrow {AC} .$

$\overrightarrow {IB}  + 2\overrightarrow {IC}  + \overrightarrow {IA}  = \overrightarrow 0 .$

$\overrightarrow {IB}  + \overrightarrow {IC}  + 2\overrightarrow {IA}  = \overrightarrow 0 .$

$2\overrightarrow {IB}  + \overrightarrow {IC}  + \overrightarrow {IA}  = \overrightarrow 0 .$

$\overrightarrow {IB}  + \overrightarrow {IC}  + \overrightarrow {IA}  = \overrightarrow 0 .$

$a.$    

$\left( {1 + \sqrt 2 } \right)a.$

$a\sqrt 5 .$

$2a\sqrt 2 .$

$\overrightarrow {AC}  + \overrightarrow {BD}  = 2\overrightarrow {BC} .$

$\overrightarrow {AC}  + \overrightarrow {BC}  = \overrightarrow {AB} .$

$\overrightarrow {AC}  - \overrightarrow {BD}  = 2\,\overrightarrow {CD} .$

$\overrightarrow {AC}  - \overrightarrow {AD}  = \overrightarrow {CD} .$

$2\vec a + \vec b,\,\,\vec a + 2\vec b.$

$2\vec a - \vec b,\,\,\vec a - 2\vec b.$

$5\vec a + \vec b,\,\, - \,10\,\vec a - 2\vec b.$

$\vec a + \vec b,\,\,\vec a - \vec b.$