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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 565
565
\[\boxed{\mathbf{565.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\ \]
\[ABCD - тетраэдр.\]
\[Доказать:\]
\[\textbf{а)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{AC}} + \overrightarrow{\text{CD}};\ \ \]
\[\textbf{б)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{DC}} + \overrightarrow{\text{AD}};\ \ \]
\[\textbf{в)}\ \overrightarrow{\text{DC}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{AC}} + \overrightarrow{\text{BA}}.\]
\[Доказательство.\]
\[\textbf{а)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{AD}}\text{\ \ }и\ \ \]
\[\overrightarrow{\text{AC}} + \overrightarrow{\text{CD}} = \overrightarrow{\text{AD}}:\]
\[\overrightarrow{\text{AB}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{AC}} + \overrightarrow{\text{CD}}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{AC}};\ \]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{AD}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{DC}} = \overrightarrow{\text{AC}}\ \]
\[(переместительный\ закон).\]
\[Следовательно:\ \]
\[\overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{DC}} + \overrightarrow{\text{AD}}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{в)}\ Переместительный\ закон:\ \]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{BD}} + \overrightarrow{\text{DC}} = \overrightarrow{\text{BC}};\]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{BD}} + \overrightarrow{\text{DC}} = \overrightarrow{\text{BC}}\ \]
\[Следовательно:\ \]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{BD}} = \overrightarrow{\text{AC}} + \overrightarrow{\text{BA}}.\]
\[Что\ и\ требовалось\ доказать.\]
