ГДЗ по геометрии 7 класс Атанасян Задание 806

\[\boxed{\mathbf{806.ОК\ ГДЗ - домашка\ на}\ 5}\]

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

\[\mathbf{Дано:}\]

\[AB - отрезок;\]

\[C \in AB;\ \ \]

\[AC\ :CB = m\ :n;\]

\[( \bullet )O - любая.\]

\[\mathbf{Доказать:}\]

\[\overrightarrow{\text{OC}} = \frac{n}{m + n}\overrightarrow{\text{OA}} + \frac{m}{m + n}\overrightarrow{\text{OB}}.\]

\[\mathbf{Доказательство.}\]

\[1)\frac{\text{AC}}{\text{CB}} = \frac{m}{n}\]

\[AC \bullet n = CB \bullet m.\ \]

\[\overrightarrow{\text{AC}} = \frac{m}{n}\overrightarrow{\text{CB}}.\]

\[2)\ По\ правилу\ треугольника:\]

\[\overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AC}}.\]

\[3)\ \overrightarrow{\text{AC}} = \frac{m}{n}\overrightarrow{\text{CB}} = \frac{m}{n}\left( \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} \right).\]

\[4)\ \overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \frac{m}{n}\left( \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} \right) =\]

\[= \overrightarrow{\text{OA}} + \frac{m}{n}\overrightarrow{\text{OB}} - \frac{m}{n}\overrightarrow{\text{OC}}\]

\[\overrightarrow{\text{OC}} + \frac{m}{n}\overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \frac{m}{n}\overrightarrow{\text{OB}}\]

\[\frac{m + n}{n}\overrightarrow{\text{OC}} = \overrightarrow{\text{OA}} + \frac{m}{n}\overrightarrow{\text{OB}}\ \ \ \ \ | \bullet \frac{n}{m + n}\]

\[\overrightarrow{\text{OC}} = \frac{n}{m + n}\overrightarrow{\text{OA}} + \frac{m}{m + n}\overrightarrow{\text{OB}}.\]

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