\[Схематический\ рисунок.\]
\[Дано:\]
\[\mathrm{\Delta}ABC;\]
\[BM - медиана;\]
\[AB = 12\ см;\]
\[BC = 14\ см;\]
\[BM = 7\ см.\]
\[Найти:\]
\[\text{AC.}\]
\[Решение.\]
\[1)\ В\ \mathrm{\Delta}ABC:\]
\[AM = CM = \frac{1}{2}AC;\]
\[AC^{2} - 24AC \bullet \cos{\angle A} = 52\]
\[24AC \bullet \cos{\angle A} = AC^{2} - 52\]
\[\cos{\angle A} = \frac{AC^{2} - 52}{24AC}.\]
\[2)\ В\ \mathrm{\Delta}ABM:\]
\[\frac{1}{4}AC^{2} - \frac{1}{2}\left( AC^{2} - 52 \right) + 95 = 0\]
\[- \frac{1}{4}AC^{2} + 26 = - 95\ \ \ \]
\[\frac{1}{4}AC^{2} = 121\]
\[AC^{2} = 484\ \ \ \]
\[AC = 22\ см.\]
\(Ответ:\ \ 22\ см.\)