\(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AE}=\frac{1}{2}.\frac{2}{3}\overrightarrow{AB}+\frac{1}{2}.\frac{1}{4}\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{8}\overrightarrow{AC}\)
\(=\frac{1}{3}\overrightarrow{AB}+\frac{1}{8}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=\frac{11}{24}\overrightarrow{AB}+\frac{1}{8}\overrightarrow{BC}\)
Đặt \(\overrightarrow{BN}=x.\overrightarrow{BC}\)
\(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}=\overrightarrow{AB}+x.\overrightarrow{BC}\)
Do A;M;N thẳng hàng \(\Rightarrow\frac{1}{\frac{11}{24}}=\frac{x}{\frac{1}{8}}\Rightarrow x=\frac{3}{11}\)
\(\Rightarrow\frac{BN}{CN}=\frac{3}{8}\)
