ГДЗ по геометрии 9 класс Атанасян ФГОС Задание 959

\[\boxed{\mathbf{959.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[Дано:\ \]

\[ABCD - параллелограмм;\ \]

\[AC \cap BD = O;\ \]

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

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

\[Выразить:\ \]

\[\overrightarrow{\text{DC}} + \overrightarrow{\text{CB}};\ \]

\[\overrightarrow{\text{BO}} + \overrightarrow{\text{OC}};\ \]

\[\overrightarrow{\text{BO}} - \overrightarrow{\text{OC}};\ \]

\[\overrightarrow{\text{BA}} - \overrightarrow{\text{DA}}.\]

\[Решение.\]

\[1)\ \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{DB}} =\]

\[= \overrightarrow{\text{AB}} - \overrightarrow{\text{AD}} =\]

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

\[2)\ \overrightarrow{\text{BO}} + \overrightarrow{\text{OC}} = \overrightarrow{\text{BC}} = \overrightarrow{\text{AD}} = \overrightarrow{b}.\]

\[3)\ \overrightarrow{\text{BO}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{BO}} - \overrightarrow{\text{AO}} =\]

\[= \overrightarrow{\text{BO}} - \left( - \overrightarrow{\text{OA}} \right) = \overrightarrow{\text{BO}} + \overrightarrow{\text{OA}} =\]

\[= \overrightarrow{\text{BA}} = - \overrightarrow{\text{AB}} =\]

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

\[4)\ \overrightarrow{\text{BA}} - \overrightarrow{\text{DA}} = \overrightarrow{\text{BA}} - \left( - \overrightarrow{\text{AD}} \right) =\]

\[= \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}} = - \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} =\]

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