Ta có:
$\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}=AB^{2}.AC^{2}$
$(\overrightarrow{AB}.\overrightarrow{AC})^{2}=(AB.AC.cos(\overrightarrow{AB},\overrightarrow{AC}))^{2}$
với $cos(\overrightarrow{AB},\overrightarrow{AC})=BAC$
=>$\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}$ = $AB^{2}.AC^{2}(1-cos^{2}BAC)$=$AB^{2}.AC^{2}(sin^{2}BAC)$
Suy ra: $\frac{1}{2}\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}$=$\frac{1}{2}AB.AC.sin^{2}BAC$=$S_{ABC}$