
\(\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{BC}=-\overrightarrow{AB}+\overrightarrow{AC}-\overrightarrow{AB}=-2\overrightarrow{AB}+\overrightarrow{AC}\). Vậy m=-2,n=1.
Gọi H là trung điểm BC, ta có: \(\cos \widehat{ACH}=\frac{HC}{AC} \Rightarrow AC=\frac{HC}{\cos 30{}^\circ } =a. \)
Do đó: \(\overrightarrow{AC}.\overrightarrow{BD}=\overrightarrow{AC}.\left(-2\overrightarrow{AB}+\overrightarrow{AC}\right)=-2\overrightarrow{AC}.\overrightarrow{AB}+\overrightarrow{AC}^{2} =-2a^{2} \cos 120{}^\circ +a^{2} =2a^{2}\) .