Bài tập 2.20 trang 92 SBT Hình học 10

 \(\begin{array}{l}
\overrightarrow {HM}  = \frac{1}{2}\left( {\overrightarrow {HB}  + \overrightarrow {HC} } \right)\\
 \Rightarrow \overrightarrow {AM} .\overrightarrow {HM}  = \frac{1}{4}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).\left( {\overrightarrow {HB}  + \overrightarrow {HC} } \right)\\
 = \frac{1}{4}\left( {\overrightarrow {AB} .\overrightarrow {HB}  + \underbrace {\overrightarrow {AB} .\overrightarrow {HC} }_{ = 0} + \underbrace {\overrightarrow {AC} .\overrightarrow {HB} }_{ = 0} + \overrightarrow {AC} .\overrightarrow {HC} } \right)\\
 = \frac{1}{4}\left( {\overrightarrow {AB} .\overrightarrow {HB}  + \overrightarrow {AC} .\overrightarrow {HC} } \right)\\
 = \frac{1}{4}\left[ {\overrightarrow {AB} \left( {\overrightarrow {HC}  + \overrightarrow {CB} } \right) + \overrightarrow {AC} .\left( {\overrightarrow {HB}  + \overrightarrow {BC} } \right)} \right]\\
 = \frac{1}{4}\left[ {\underbrace {\overrightarrow {AB} .\overrightarrow {HC} }_{ = 0} + \overrightarrow {AB} .\overrightarrow {CB}  + \underbrace {\overrightarrow {AC} .\overrightarrow {HB} }_{ = 0} + \overrightarrow {AC} .\overrightarrow {BC} } \right]\\
 = \frac{1}{4}\left( {\overrightarrow {AB} .\overrightarrow {HC}  + \overrightarrow {AC} .\overrightarrow {BC} } \right) = \frac{1}{4}\left( {\overrightarrow {AB} .\overrightarrow {CB}  - \overrightarrow {AC} .\overrightarrow {CB} } \right)\\
 = \frac{1}{4}\overrightarrow {CB} .\left( {\underbrace {\overrightarrow {AB}  - \overrightarrow {AC} }_{\overrightarrow {CB} }} \right) = \frac{1}{4}{\overrightarrow {CB} ^2} = \frac{1}{4}{\overrightarrow {BC} ^2}
\end{array}\)

-- Mod Toán 10 HỌC247