Theo Pytago: \(BC^2=AB^2+AC^2\\ \Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{10^2-6^2}=8\left(cm\right)\)
△ABC có BM là phân giác
\(\Rightarrow\dfrac{BA}{BC}=\dfrac{MA}{MC}\\ \Rightarrow\dfrac{MA}{MC}=\dfrac{6}{10}=\dfrac{3}{5}\\ \Rightarrow\dfrac{MA}{3}=\dfrac{MC}{5}=\dfrac{MA+MC}{3+5}=\dfrac{8}{8}=1\\ \Rightarrow AM=1\cdot3=3\left(cm\right)\)