ГДЗ по геометрии 9 класс Атанасян Задание 805

\[\boxed{\mathbf{805.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\mathbf{\ задачи:}\]

\[\mathbf{Дано:}\]

\[A;B;\ C - произвольные\ точки;\]

\[\overrightarrow{\text{BC}} = \frac{1}{2}\overrightarrow{\text{AB}};\]

\[точка\ O - любая.\]

\[\mathbf{Доказать:}\]

\[\overrightarrow{\text{OB}} = \frac{1}{3}\overrightarrow{\text{OA}} + \frac{2}{3}\overrightarrow{\text{OC}}.\]

\[\mathbf{Доказательство.}\]

\[1)\ По\ правилу\ треугольника:\]

\[\overrightarrow{\text{OB}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}}.\]

\[2)\ \overrightarrow{\text{BC}} = \frac{1}{2}\overrightarrow{\text{AB}}:\]

\[\overrightarrow{\text{AB}} = 2\overrightarrow{\text{BC}} = 2 \bullet \left( \overrightarrow{\text{OC}} - \overrightarrow{\text{OB}} \right)\text{.\ }\]

\[3)\ \overrightarrow{\text{OB}} = \overrightarrow{\text{OA}} + 2\overrightarrow{\text{OC}} - 2\overrightarrow{\text{OB}}\]

\[3\overrightarrow{\text{OB}} = \overrightarrow{\text{OA}} + 2\overrightarrow{\text{OC}}\ \ \ \ \ \ |\ :3\]

\[\overrightarrow{\text{OB}} = \frac{1}{3}\overrightarrow{\text{OA}} + \frac{2}{3}\overrightarrow{\text{OC}}.\]

\[\mathbf{Что\ и\ требовалось\ доказать.}\]