ГДЗ по геометрии 8 класс Атанасян Задание 768

\[\boxed{\mathbf{768.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[Дано:\ \]

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

\[AM = MB;\]

\[AN = NC;\ \]

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

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

\[Выразить:\]

\[\overrightarrow{\text{BM}};\ \overrightarrow{\text{NC}};\ \overrightarrow{\text{MN}};\ \overrightarrow{\text{BN}}.\]

\[Решение.\]

\[1)\ \overrightarrow{\text{BM}} = \overrightarrow{\text{MA}} = - \overrightarrow{a}.\]

\[2)\ \overrightarrow{\text{NC}} = \overrightarrow{\text{AN}} = \overrightarrow{b}.\]

\[3)\ \overrightarrow{\text{MN}} = \overrightarrow{\text{AN}} - \overrightarrow{\text{AM}} =\]

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

\[4)\ \overrightarrow{\text{BN}} = \overrightarrow{\text{BM}} + \overrightarrow{\text{MN}} =\]

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

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

\[(по\ правилу\ треугольника).\]