Cho tam giác ABC, trực tâm H, M là trung điểm BC. Chứng minh vecto MH. vecto MA
Cho tam giác ABC, trực tâm H, M là trung điểm BC. Chứng minh \(\overrightarrow {MH} .\overrightarrow {MA} = \frac{1}{4}B{C^2}\).
Vì M là trung điểm BC nên: \(\left\{ \begin{array}{l}\overrightarrow {AM} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\\\overrightarrow {HM} = \frac{1}{2}\left( {\overrightarrow {HB} + \overrightarrow {HC} } \right)\end{array} \right.\)
Có: \(\overrightarrow {MH} .\overrightarrow {MA} = \overrightarrow {AM} .\overrightarrow {HM} \)
\( = \frac{1}{4}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\left( {\overrightarrow {HB} + \overrightarrow {HC} } \right)\)
\( = \frac{1}{4}\left( {\overrightarrow {AB} .\overrightarrow {HB} + \overrightarrow {AC} .\overrightarrow {HB} + \overrightarrow {AB} .\overrightarrow {HC} + \overrightarrow {AC} .\overrightarrow {HC} } \right)\)(do AC vuông góc HB nên \[\overrightarrow {AC} .\overrightarrow {HB} = \overrightarrow {AB} .\overrightarrow {HC} = \overrightarrow 0 \])
\( = \frac{1}{4}\left( {\overrightarrow {AB} .\overrightarrow {HB} + \overrightarrow {AC} .\overrightarrow {HC} } \right)\)
\( = \frac{1}{4}\left( {\overrightarrow {HB} \left( {\overrightarrow {AC} - \overrightarrow {BC} } \right) + \overrightarrow {HC} \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right)} \right)\)
\( = \frac{1}{4}\left( {\overrightarrow {HB} .\overrightarrow {AC} - \overrightarrow {HB} .\overrightarrow {BC} + \overrightarrow {HC} .\overrightarrow {AB} + \overrightarrow {HC} .\overrightarrow {BC} } \right)\)
\( = \frac{1}{4}\left( { - \overrightarrow {HB} .\overrightarrow {BC} + \overrightarrow {HC} .\overrightarrow {BC} } \right)\)
\( = \frac{1}{4}\overrightarrow {BC} \left( { - \overrightarrow {HB} + \overrightarrow {HC} } \right)\)
\( = \frac{1}{4}\overrightarrow {BC} .\overrightarrow {BC} = \frac{1}{4}B{C^2}\).