\[\boxed{\mathbf{805.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[A;B;\ C - произвольные\ точки;\]
\[\overrightarrow{\text{BC}} = \frac{1}{2}\overrightarrow{\text{AB}};\]
\[точка\ O - любая.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{OB}} = \frac{1}{3}\overrightarrow{\text{OA}} + \frac{2}{3}\overrightarrow{\text{OC}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{OB}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}}.\]
\[2)\ \overrightarrow{\text{BC}} = \frac{1}{2}\overrightarrow{\text{AB}}:\]
\[\overrightarrow{\text{AB}} = 2\overrightarrow{\text{BC}} = 2 \bullet \left( \overrightarrow{\text{OC}} - \overrightarrow{\text{OB}} \right)\text{.\ }\]
\[3)\ \overrightarrow{\text{OB}} = \overrightarrow{\text{OA}} + 2\overrightarrow{\text{OC}} - 2\overrightarrow{\text{OB}}\]
\[3\overrightarrow{\text{OB}} = \overrightarrow{\text{OA}} + 2\overrightarrow{\text{OC}}\ \ \ \ \ \ |\ :3\]
\[\overrightarrow{\text{OB}} = \frac{1}{3}\overrightarrow{\text{OA}} + \frac{2}{3}\overrightarrow{\text{OC}}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]