\[\boxed{\mathbf{769.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\ \]
\[BB_{1} - медиана;\ \]
\[\overrightarrow{x} = \overrightarrow{AB_{1}};\ \]
\[\overrightarrow{y} = \overrightarrow{\text{AB}}.\]
\[Выразить:\]
\[\overrightarrow{B_{1}C};\ \overrightarrow{BB_{1}};\ \overrightarrow{\text{BA}};\ \overrightarrow{\text{BC}}.\]
\[Решение.\]
\[1)\ \overrightarrow{B_{1}C} = \overrightarrow{AB_{1}} = \overrightarrow{x}.\]
\[2)\ \overrightarrow{BB_{1}} = \overrightarrow{AB_{1}} - \overrightarrow{\text{AB}} =\]
\[= \overrightarrow{x} - \overrightarrow{y}\ (по\ правилу\ треугольника).\]
\[3)\ \overrightarrow{\text{BA}} = - \overrightarrow{\text{AB}} = - \overrightarrow{y}.\]
\[4)\ \overrightarrow{\text{BC}} = \overrightarrow{\text{AC}} - \overrightarrow{\text{AB}} =\]
\[= \overrightarrow{AB_{1}} + \overrightarrow{B_{1}C} - \overrightarrow{\text{AB}} =\]
\[= \overrightarrow{x} + \overrightarrow{x} - \overrightarrow{y} = 2\overrightarrow{x} - \overrightarrow{y}\]
\((по\ правилу\ треугольника).\)