Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\overrightarrow {BC} .\overrightarrow {MH} \\
= 2.\overrightarrow {BC} .\left( {\overrightarrow {MA} + \overrightarrow {AH} } \right)\\
= 2\overrightarrow {BC} .\overrightarrow {MA} + 2\overrightarrow {BC} .\overrightarrow {AH} \\
= \overrightarrow {BC} .2\overrightarrow {MA} \,\,\,\,\,\,\left( {AH \bot BC \Rightarrow \overrightarrow {BC} .\overrightarrow {AH} = 0} \right)\\
= \left( {\overrightarrow {BA} + \overrightarrow {AC} } \right).\left( {\overrightarrow {MB} + \overrightarrow {BA} + \overrightarrow {MC} + \overrightarrow {CA} } \right)\\
= \left( {\overrightarrow {BA} + \overrightarrow {AC} } \right).\left( {\overrightarrow {BA} + \overrightarrow {CA} } \right)\,\,\,\,\,\left( {\overrightarrow {MB} + \overrightarrow {MC} = \overrightarrow 0 } \right)\\
= {\overrightarrow {BA} ^2} + \overrightarrow {BA} .\overrightarrow {CA} + \overrightarrow {AC} .\overrightarrow {BA} - {\overrightarrow {AC} ^2}\\
= A{B^2} + \overrightarrow {BA} .\left( {\overrightarrow {CA} + \overrightarrow {AC} } \right) - {\overrightarrow {AC} ^2}\\
= A{B^2} - A{C^2}
\end{array}\)
