Решебник по геометрии 11 класс. Атанасян ФГОС 585

\[\boxed{\mathbf{585.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[Доказать:\]

\[\overrightarrow{AC_{1}} + \overrightarrow{B_{1}D} = 2\overrightarrow{\text{BC}}.\]

\[Доказательство.\]

\[1)\ ABCDA_{1}B_{1}C_{1}D_{1} -\]

\[параллелепипед.\]

\[AD = BC;\ AD \parallel BC:\]

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

\[B_{1}A = C_{1}D;\ \ B_{1}A \parallel C_{1}D:\]

\(\overrightarrow{B_{1}A} = \overrightarrow{C_{1}D} = - \overrightarrow{DC_{1}}.\)

\[2)\ \overrightarrow{B_{1}D} = \overrightarrow{B_{1}A} + \overrightarrow{\text{AD}};\ \ \ \ \]

\[\overrightarrow{AC_{1}} = \overrightarrow{\text{AD}} + \overrightarrow{DC_{1}}:\]

\[\overrightarrow{AC_{1}} + \overrightarrow{B_{1}D} =\]

\[= \overrightarrow{\text{AD}} + \overrightarrow{DC_{1}} + \overrightarrow{B_{1}A} + \overrightarrow{\text{AD}} =\]

\[= 2\overrightarrow{\text{AD}} + \overrightarrow{DC_{1}} + \overrightarrow{B_{1}A} =\]

\[= 2\overrightarrow{\text{BC}} + \overrightarrow{DC_{1}} - \overrightarrow{DC_{1}} = 2\overrightarrow{\text{BC}}.\]

\[Что\ и\ требовалось\ доказать.\]