Giải thích các bước giải:
Ta có :
$\dfrac{DB}{DC}=\dfrac{AB}{AC}=2\to \dfrac{DB}{DB+DC}=\dfrac{2}{2+1}\to \dfrac{DB}{BC}=\dfrac{2}3\to \vec{BD}=\dfrac{2}3\vec{BC}$
$\to\vec{AD}=\vec{AB}+\vec{BD}=\vec{AB}+\dfrac{2}3\vec{BC}=\vec{AB}+\dfrac{2}3(\vec{AC}-\vec{AB})=\dfrac{1}3\vec{AB}+\dfrac{2}3\vec{AC}$
$\to m=\dfrac{1}3, k=\dfrac{2}3\to S=1347$