Ta chứng minh: \(\overrightarrow {AM} \,.\,\overrightarrow {DB} = 0\)
Ta có: \[\overrightarrow {BD} = \overrightarrow {BH} + \overrightarrow {HD} = \overrightarrow {HC} + \overrightarrow {HD} \]
\(\overrightarrow {AM} = \frac{1}{2}\left( {\overrightarrow {AH} + \overrightarrow {AD} } \right)\)
Do đó: \(\overrightarrow {AM} \,.\,\overrightarrow {DB} = \frac{1}{2}\left( {\overrightarrow {AH} + \overrightarrow {AD} } \right)\left( {\overrightarrow {HC} + \overrightarrow {HD} } \right)\)
\( = \frac{1}{2}\left( {\overrightarrow {AH} \,.\,\overrightarrow {HC} + \overrightarrow {AH} \,.\,\overrightarrow {HD} + \overrightarrow {AD} \,.\,\overrightarrow {HC} + \overrightarrow {AD} \,.\,\overrightarrow {HD} } \right)\)
Mà: \(\left\{ \begin{array}{l}\overrightarrow {AH} \,.\,\overrightarrow {HC} = 0\;\left( {do\;AH \bot BC} \right)\\\overrightarrow {AD} \,.\,\overrightarrow {HD} = 0\;\left( {do\;AC \bot HD} \right)\end{array} \right.\)
Do đó: \(\overrightarrow {AM} \,.\,\overrightarrow {DB} = \frac{1}{2}\left( {\overrightarrow {AH} \,.\,\overrightarrow {HD} + \overrightarrow {AD} \,.\,\overrightarrow {HC} } \right)\)
\( = \frac{1}{2}\left[ {\overrightarrow {AH} \,.\,\overrightarrow {HD} + \left( {\overrightarrow {AH} + \overrightarrow {HD} } \right)\,.\,\overrightarrow {HC} } \right]\)
\( = \frac{1}{2}\left( {\overrightarrow {AH} \,.\,\overrightarrow {HD} + \overrightarrow {HD} \,.\,\overrightarrow {HC} } \right)\)vì\(\overrightarrow {AH} \,.\,\overrightarrow {HC} = 0\)
\(\overrightarrow {AM} \,.\,\overrightarrow {DB} = \frac{1}{2}\overrightarrow {HD} \left( {\overrightarrow {AH} + \,\overrightarrow {HC} } \right) = \frac{1}{2}\overrightarrow {HD} \,.\,\overrightarrow {AC} = 0\) (vì AC ^ HD)
Vậy AM ^ DB.