Решебник по геометрии 10 класс Атанасян ФГОС 614

\[\boxed{\mathbf{614.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Рисунок\ к\ задаче:173.\]

\[ABCDEF - правильный\ \]

\[октаэдр,\ все\ его\ ребра\ равны.\]

\[\textbf{а)}\ ABFD - ромб\ (по\ правилу\ \]

\[параллелограмма):\]

\[\overrightarrow{\text{AB}} + \overrightarrow{\text{FB}} = - \overrightarrow{\text{BA}} - \overrightarrow{\text{BF}} =\]

\[= - \left( \overrightarrow{\text{BA}} + \overrightarrow{\text{BF}} \right) = - \overrightarrow{\text{BD}} = \overrightarrow{\text{DB}}.\]

\[\textbf{б)}\ ACFE - ромб\ (по\ правилу\ \]

\[параллелограмма):\]

\[\overrightarrow{\text{AC}} - \overrightarrow{\text{CF}} = - \overrightarrow{\text{CA}} - \overrightarrow{\text{CF}} =\]

\[= - \left( \overrightarrow{\text{CA}} + \overrightarrow{\text{CF}} \right) = - \overrightarrow{\text{CE}} = \overrightarrow{\text{EC}}.\]

\[\textbf{в)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{AE}} =\]

\[= \left( \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} \right) + \left( \overrightarrow{\text{AC}} + \overrightarrow{\text{AE}} \right) =\]

\[= \overrightarrow{\text{AF}} + \overrightarrow{\text{AF}} = 2\overrightarrow{\text{AF}}.\]

\[Что\ и\ требовалось\ доказать.\]