ГДЗ по геометрии 9 класс Атанасян Задание 787

\[\boxed{\mathbf{787.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]

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

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}DEF;\]

\[EG - медиана;\]

\[\text{EO} = \text{OG};\]

\[\overrightarrow{a} = \overrightarrow{\text{ED}};\ \]

\[\overrightarrow{b} = \overrightarrow{\text{EF}}.\]

\[Выразить:\]

\(\overrightarrow{\text{DO}}\ через\ \overrightarrow{a}\text{\ \ }и\ \overrightarrow{b}\).

\[\mathbf{Решение.}\]

\[1)\ \overrightarrow{\text{ED}} + \overrightarrow{\text{DF}} = \overrightarrow{\text{EF}}\]

\[\overrightarrow{\text{DF}} = \overrightarrow{\text{EF}} - \overrightarrow{\text{ED}} = \overrightarrow{b} - \overrightarrow{a}.\]

\[2)\ \overrightarrow{\text{GE}} = \overrightarrow{\text{GD}} + \overrightarrow{\text{DE}} =\]

\[= - \frac{1}{2}\overrightarrow{\text{DF}} - \overrightarrow{\text{ED}} =\]

\[= - \frac{1}{2} \bullet \left( \overrightarrow{b} - \overrightarrow{a} \right) - \overrightarrow{a} =\]

\[= - \frac{1}{2}\overrightarrow{b} + \frac{1}{2}\overrightarrow{a} - \overrightarrow{a} = - \frac{1}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a}.\]

\[3)\ \overrightarrow{\text{DO}} = \overrightarrow{\text{DG}} + \overrightarrow{\text{GO}} =\]

\[= \frac{1}{2}\overrightarrow{\text{DF}} + \frac{1}{2}\overrightarrow{\text{GE}} =\]

\[= \frac{1}{2} \bullet \left( \overrightarrow{b} - \overrightarrow{a} \right) + \frac{1}{2} \bullet \left( - \frac{1}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a} \right) =\]

\[= \frac{1}{2}\overrightarrow{b} - \frac{1}{2}\overrightarrow{a} - \frac{1}{4}\overrightarrow{b} - \frac{1}{4}\overrightarrow{a} =\]

\[= \frac{1}{4}\overrightarrow{b} - \frac{3}{4}\ \overrightarrow{a}.\]

\[Ответ:\ \overrightarrow{\text{DO}} = \frac{1}{4}\overrightarrow{b} - \frac{3}{4}\ \overrightarrow{a}.\]