Trước hết, theo các tính chất về trung tuyến ta có:
\(\overrightarrow{CN}=\frac{1}{2}\overrightarrow{CD}+\frac{1}{2}\overrightarrow{CO}\) ;
\(\overrightarrow{CI}=\frac{1}{2}\overrightarrow{CM}+\frac{1}{2}\overrightarrow{CN}=\frac{1}{4}\overrightarrow{CB}+\frac{1}{2}\left(\frac{1}{2}\overrightarrow{CD}+\frac{1}{2}\overrightarrow{CO}\right)=\frac{1}{4}\left(\overrightarrow{CB}+\overrightarrow{CD}\right)+\frac{1}{4}\overrightarrow{CO}\)
\(\Rightarrow\overrightarrow{CI}=\frac{1}{4}.2\overrightarrow{CO}+\frac{1}{4}\overrightarrow{CO}=\frac{3}{4}\overrightarrow{CO}\)
\(\Rightarrow C;I;A\) thẳng hàng và \(\overrightarrow{AI}=\frac{5}{8}\overrightarrow{AC}\)
\(\overrightarrow{BI}=\overrightarrow{BA}+\overrightarrow{AI}=\overrightarrow{BA}+\frac{5}{8}\overrightarrow{AC}=\overrightarrow{BA}+\frac{5}{8}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\)
\(\Rightarrow\overrightarrow{BI}=\frac{3}{8}\overrightarrow{BA}+\frac{5}{8}\overrightarrow{AD}\)
\(\overrightarrow{BI}=\frac{1}{2}\left(\overrightarrow{BM}+\overrightarrow{BN}\right)=\frac{1}{2}\cdot\frac{3}{4}\cdot\overrightarrow{BD}+\frac{1}{4}\overrightarrow{BC}\)
\(\overrightarrow{BI}=\frac{3}{8}.\frac{\overrightarrow{BA}+\overrightarrow{BC}}{2}+\frac{1}{4}\overrightarrow{BC}=\frac{3}{16}\overrightarrow{BA}+\frac{7}{16}\overrightarrow{BC}\)
Có sai mong bạn thông cảm ạ
