Ta có: \(AB\perp AC\left(\Delta ABC-vuông-tại-A\right)\)
Và: \(MN\perp AC\left(gt\right)\)
\(\Rightarrow AB//MN\)
\(\Rightarrow\Delta CMN~\Delta CBA\)
\(\Rightarrow\frac{CN}{CA}=\frac{CM}{CB}\left(1\right)\)
Ta có: \(BC^2=AB^2+AC^2\left(Định-li-pitago\right)\)
\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}\)
\(\Rightarrow BC=10cm\)
Từ: \(\left(1\right)\Rightarrow\frac{CN}{8}=\frac{3}{10}\Rightarrow CN=\frac{3.8}{10}=2,4cm\)