ГДЗ по геометрии 9 класс Атанасян ФГОС Задание 974

\[\boxed{\mathbf{974}\mathbf{.еуроки - ответы\ на\ пятёрку}}\]

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

\[\mathbf{Дано:}\]

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

\[AA_{1};BB_{1};CC_{1} - медианы.\]

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

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

\[Выразить:\]

\[\overrightarrow{AA_{1}};\ \ \overrightarrow{BB_{1}\ };\ \overrightarrow{CC_{1}}\mathbf{.}\]

\[\mathbf{Решение.}\]

\[1)\ \overrightarrow{BB_{1}} = \overrightarrow{\text{BA}} + \ \overrightarrow{AB_{1}} =\]

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

\[2)\ \overrightarrow{CC_{1}} = \overrightarrow{\text{CB}} + \overrightarrow{BC_{1}} =\]

\[= \overrightarrow{\text{CA}} + \overrightarrow{AC_{1}} = - \overrightarrow{\text{AC}} + \frac{1}{2}\overrightarrow{\text{AB}} =\]

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

\[3)\ \overrightarrow{\text{BC}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AC}} =\]

\[= - \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} = - \overrightarrow{b} + \overrightarrow{a}.\]

\[4)\ \overrightarrow{AA_{1}} = \overrightarrow{\text{AB}} + \overrightarrow{BA_{1}} =\]

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

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

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