Решебник по геометрии 7 класс Атанасян ФГОС Задание 953

 

\[\boxed{\mathbf{953.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Решение\ задачи\ дано\ в\ }\]

\[\mathbf{учебнике.}\]

\[\boxed{\mathbf{953.еуроки - ответы\ на\ пятёрку}}\]

\[Дано:\]

\[X,Y,Z - произвольные\ точки.\]

\[Доказать:\ \]

\[\overrightarrow{p} = \overrightarrow{\text{XY}} + \overrightarrow{\text{ZX}} + \overrightarrow{\text{YZ}} = \overrightarrow{0};\]

\[\overrightarrow{q} = \left( \overrightarrow{\text{XY}} - \overrightarrow{\text{XZ}} \right) + \overrightarrow{\text{YZ}} = \overrightarrow{0};\]

\[\overrightarrow{z} = \left( \overrightarrow{\text{ZY}} - \overrightarrow{\text{XY}} \right) - \overrightarrow{\text{ZX}} = \overrightarrow{0}.\]

\[Доказательство.\]

\[Воспользуемся\ правилом\ \]

\[многоугольника.\]

\[1)\ \overrightarrow{p} = \overrightarrow{\text{XY}} + \overrightarrow{\text{ZX}} + \overrightarrow{\text{YZ}} =\]

\[= \overrightarrow{\text{XY}} + \overrightarrow{\text{YZ}} + \overrightarrow{\text{ZX}} = \overrightarrow{\text{XX}} = \overrightarrow{0}.\]

\[2)\ \overrightarrow{q} = \left( \overrightarrow{\text{XY}} - \overrightarrow{\text{XZ}} \right) + \overrightarrow{\text{YZ}} =\]

\[= \left( \overrightarrow{\text{XY}} + \overrightarrow{\text{ZX}} \right) + \overrightarrow{\text{YZ}} =\]

\[= \left( \overrightarrow{\text{XY}} + \overrightarrow{\text{YZ}} \right) + \overrightarrow{\text{ZX}} =\]

\[= \overrightarrow{\text{XZ}} + \overrightarrow{\text{ZX}} = \overrightarrow{\text{XX}} = \overrightarrow{0}.\]

\[3)\ \overrightarrow{z} = \left( \overrightarrow{\text{ZY}} - \overrightarrow{\text{XY}} \right) - \overrightarrow{\text{ZX}} =\]

\[= \left( \overrightarrow{\text{ZY}} + \overrightarrow{\text{YX}} \right) + \overrightarrow{\text{XZ}} =\]

\[= \overrightarrow{\text{ZX}} + \overrightarrow{\text{XZ}} = \overrightarrow{\text{ZZ}} = \overrightarrow{0}.\]

\[Таким\ образом:\ \]

\[\overrightarrow{p} = \overrightarrow{q} = \overrightarrow{z} = \overrightarrow{0}.\]

\[Что\ и\ требовалось\ доказать\text{.\ }\]