ГДЗ по геометрии 9 класс Атанасян Задание 770

\[\boxed{\mathbf{770.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[Дано:\ \]

\[ABCD - параллелограмм;\]

\[\textbf{а)}\ \overrightarrow{a} = \overrightarrow{\text{AB}};\ \overrightarrow{b} = \overrightarrow{\text{BC}};\]

\[\textbf{б)}\ \overrightarrow{a} = \overrightarrow{\text{CB}};\ \overrightarrow{b} = \overrightarrow{\text{CD}};\]

\[\textbf{в)}\ \overrightarrow{a} = \overrightarrow{\text{AB}};\ \overrightarrow{b} = \overrightarrow{\text{DA}}.\]

\[Выразить:\ \]

\[\overrightarrow{\text{AC}}.\]

\[Решение.\]

\[\textbf{а)}\ \overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} =\]

\[= \overrightarrow{a} + \overrightarrow{b}\ (по\ правилу\ треугольника).\]

\[\textbf{б)}\ \overrightarrow{\text{AC}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{DC}} =\]

\[= \overrightarrow{\text{BC}} + \left( - \overrightarrow{\text{CD}} \right) =\]

\[= - \overrightarrow{\text{CB}} + \left( - \overrightarrow{\text{CD}} \right) =\]

\[= - \overrightarrow{a} + \left( - \overrightarrow{b} \right) = - \overrightarrow{a} - \overrightarrow{b}.\]

\[\textbf{в)}\ \overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} =\]

\[= \overrightarrow{\text{AB}} + \left( - \overrightarrow{\text{DA}} \right) = \overrightarrow{a} + \left( - \overrightarrow{b} \right) =\]

\[= \overrightarrow{a} - \overrightarrow{b}\ (по\ правилу\ \]

\[треугольника).\]