\[\boxed{\mathbf{806.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[AB - отрезок;\]
\[C \in AB;\ \ \]
\[AC\ :CB = m\ :n;\]
\[( \bullet )O - любая.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{OC}} = \frac{n}{m + n}\overrightarrow{\text{OA}} + \frac{m}{m + n}\overrightarrow{\text{OB}}.\]
\[\mathbf{Доказательство.}\]
\[1)\frac{\text{AC}}{\text{CB}} = \frac{m}{n}\]
\[AC \bullet n = CB \bullet m.\ \]
\[\overrightarrow{\text{AC}} = \frac{m}{n}\overrightarrow{\text{CB}}.\]
\[2)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AC}}.\]
\[3)\ \overrightarrow{\text{AC}} = \frac{m}{n}\overrightarrow{\text{CB}} = \frac{m}{n}\left( \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} \right).\]
\[4)\ \overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \frac{m}{n}\left( \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} \right) =\]
\[= \overrightarrow{\text{OA}} + \frac{m}{n}\overrightarrow{\text{OB}} - \frac{m}{n}\overrightarrow{\text{OC}}\]
\[\overrightarrow{\text{OC}} + \frac{m}{n}\overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \frac{m}{n}\overrightarrow{\text{OB}}\]
\[\frac{m + n}{n}\overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \frac{m}{n}\overrightarrow{\text{OB}}\ \ \ \ \ | \bullet \frac{n}{m + n}\]
\[\overrightarrow{\text{OC}} = \frac{n}{m + n}\overrightarrow{\text{OA}} + \frac{m}{m + n}\overrightarrow{\text{OB}}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]