Sử dụng định lý con nhím để chứng minh
Ta có
$\begin{array}{l}
\overrightarrow {CM} = \dfrac{{AM}}{{AB}}.\overrightarrow {CB} + \dfrac{{BM}}{{AB}}.\overrightarrow {CA} \\
\Leftrightarrow AB.\overrightarrow {CM} = AM.\overrightarrow {CB} + BM.\overrightarrow {CA} \\
\Leftrightarrow A{B^2}.C{M^2} = A{M^2}.C{B^2} + B{M^2}.C{A^2} + 2AM.BM.\overrightarrow {CB} .\overrightarrow {CA} \\
\Leftrightarrow A{B^2}.C{M^2} = A{M^2}.C{B^2} + B{M^2}.C{A^2} + 2AM.BM.CB.CA.\cos \left( {\overrightarrow {CB} ,\overrightarrow {CA} } \right)\\
\Leftrightarrow A{B^2}.C{M^2} = A{M^2}.C{B^2} + B{M^2}.C{A^2} + \dfrac{{C{B^2} + C{A^2} - A{B^2}}}{{2CB.CA}}.2CB.CA.AM.BM\\
\Leftrightarrow A{B^2}.C{M^2} = A{M^2}.C{B^2} + B{M^2}.C{A^2} + AM.BM\left( {{a^2} + {b^2} - {c^2}} \right)\\
\Leftrightarrow {c^2}C{M^2} = A{M^2}{a^2} + B{M^2}{b^2} + AM.BM({a^2} + {b^2} - {c^2})\\
\Leftrightarrow A{M^2} = {b^2}B{M^2} + {c^2}C{M^2} + \left( {{b^2} + {c^2} - {a^2}} \right)BM.CM
\end{array}$
