\(\begin{array}{l}
Ta\,\,\,co:\,\,\,BC = \sqrt {A{C^2} + A{B^2}} = \sqrt {{{10}^2} + {{12}^2}} = 2\sqrt {61} .\\
\Rightarrow CH = \frac{{A{C^2}}}{{BC}} = \frac{{{{10}^2}}}{{2\sqrt {61} }} = \frac{{50\sqrt {61} }}{{61}}.\\
Co\,\,\,\,\left\{ \begin{array}{l}
AD = 2DB\\
AD + DB = AB = 12
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
AD = 8\\
BD = 4
\end{array} \right..\\
\Rightarrow \tan \angle B = \frac{{AC}}{{AB}} = \frac{{10}}{{12}} = \frac{5}{6} \Rightarrow \angle B \approx {39^0}48'\\
\tan \angle ADC = \frac{{AC}}{{AD}} = \frac{{10}}{8} = \frac{5}{4} \Rightarrow \angle D \approx {51^0}20'\\
\Rightarrow \angle CDB = {180^0} - \angle ADC = {128^0}39'\\
\Rightarrow \angle DCB = {180^0} - \angle CDB - \angle B = {180^0} - {128^0}39' - {39^0}48' = {11^0}33'\\
\Rightarrow \cos \angle DCB = \cos \angle DMH = \frac{{CH}}{{CM}}\\
\Leftrightarrow CM = \frac{{CH}}{{\cos \angle DMH}} = \frac{{50\sqrt {61} }}{{61.\cos {{11}^0}33'}} \approx 6,69.
\end{array}\)
