Bài tập 9 trang 52 SGK Hình học 10 NC

Ta có:

\(\begin{array}{l}
\overrightarrow {AD}  = \frac{1}{2}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right)\\
\overrightarrow {BE}  = \frac{1}{2}\left( {\overrightarrow {BA}  + \overrightarrow {BC} } \right)\\
\overrightarrow {CF}  = \frac{1}{2}\left( {\overrightarrow {CA}  + \overrightarrow {CB} } \right)
\end{array}\)

Do đó

\(\begin{array}{*{20}{l}}
{\overrightarrow {BC} .\overrightarrow {AD}  + \overrightarrow {CA} .\overrightarrow {BE}  + \overrightarrow {AB} .\overrightarrow {CF} }\\
\begin{array}{l}
 = \frac{1}{2}\overrightarrow {BC} \left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right) + \frac{1}{2}\overrightarrow {CA} \left( {\overrightarrow {BA}  + \overrightarrow {BC} } \right)\\
 + \frac{1}{2}\overrightarrow {AB} \left( {\overrightarrow {CA}  + \overrightarrow {CB} } \right)
\end{array}\\
{ = \frac{1}{2}\left( \begin{array}{l}
\overrightarrow {BC} .\overrightarrow {AB}  + \overrightarrow {BC} .\overrightarrow {AC}  + \overrightarrow {CA} .\overrightarrow {BA} \\
 + \overrightarrow {CA} .\overrightarrow {BC}  + \overrightarrow {AB} .\overrightarrow {CA}  + \overrightarrow {AB} .\overrightarrow {CB} 
\end{array} \right)}\\
\begin{array}{l}
 = \frac{1}{2}\left( {\overrightarrow {BC} .\overrightarrow {AB}  + \overrightarrow {AB} .\overrightarrow {CB} } \right) + \frac{1}{2}\left( {\overrightarrow {BC} .\overrightarrow {AC}  + \overrightarrow {CA} .\overrightarrow {BC} } \right)\\
 + \frac{1}{2}\left( {\overrightarrow {CA} .\overrightarrow {BA}  + \overrightarrow {AB} .\overrightarrow {CA} } \right) = 0
\end{array}
\end{array}\)

-- Mod Toán 10 HỌC247