\(AC=\sqrt{BC^2-AB^2}=a\)
Gọi M là trung điểm AC \(\Rightarrow AM=\dfrac{1}{2}AC=\dfrac{a}{2}\)
\(\Rightarrow BM=\sqrt{AM^2+AB^2}=\dfrac{a\sqrt{13}}{2}\)
\(\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=\left|2\overrightarrow{BM}\right|=2BM=a\sqrt{13}\)
