Решебник по геометрии 10 класс Атанасян ФГОС 575

\[\boxed{\mathbf{575.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} -\]

\[параллелепипед;\]

\[O - произвольная\ очка.\]

\[Доказать:\]

\[\overrightarrow{\text{OA}} + \overrightarrow{OC_{1}} = \overrightarrow{\text{OC}} + \overrightarrow{OA_{1}}.\]

\[Доказательство.\]

\[1)\ \overrightarrow{\text{OA}} = \overrightarrow{OA_{1}} + \overrightarrow{A_{1}A}:\ \]

\[\overrightarrow{A_{1}A} = \overrightarrow{\text{OA}} - \overrightarrow{OA_{1}}.\]

\[2)\ \overrightarrow{\text{OC}} = \overrightarrow{OC_{1}} + \overrightarrow{C_{1}C}:\ \]

\[\overrightarrow{C_{1}C} = \overrightarrow{\text{OC}} - \overrightarrow{OC_{1}}.\]

\[3)\ A_{1}A \parallel C_{1}C;\text{\ \ }A_{1}A = C_{1}C:\]

\[\overrightarrow{A_{1}A} = \overrightarrow{C_{1}C}.\]

\[4)\ Получаем:\]

\[\overrightarrow{\text{OA}} - \overrightarrow{OA_{1}} = \overrightarrow{\text{OC}} - \overrightarrow{OC_{1}};\]

\[\overrightarrow{\text{OA}} + \overrightarrow{OC_{1}} = \overrightarrow{\text{OC}} + \overrightarrow{OA_{1}}\text{.\ }\]

\[Что\ и\ требовалось\ доказать.\]