Решебник по геометрии 11 класс. Атанасян ФГОС 568

\[\boxed{\mathbf{568.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCD - параллелограмм;\ \]

\[O - произвольная\ точка.\]

\[Доказать:\]

\[\textbf{а)}\ \overrightarrow{\text{OB}} - \overrightarrow{\text{OA}} = \overrightarrow{\text{OC}} - \overrightarrow{\text{OD}};\ \ \]

\[\textbf{б)}\ \overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{DA}}.\]

\[Доказательство.\]

\[\textbf{а)}\ По\ правилу\ треугольника:\]

\[\overrightarrow{\text{OB}} + \overrightarrow{\text{BA}} = \overrightarrow{\text{OA}};\ \ \]

\[\overrightarrow{\text{OC}} + \overrightarrow{\text{CD}} = \overrightarrow{\text{OD}}.\]

\[Отсюда:\]

\[\overrightarrow{\text{OB}} - \overrightarrow{\text{OA}} = \overrightarrow{\text{OB}} - \left( \overrightarrow{\text{OB}} + \overrightarrow{\text{BA}} \right) =\]

\[= - \overrightarrow{\text{BA}} = \overrightarrow{\text{AB}};\]

\[\overrightarrow{\text{OC}} - \overrightarrow{\text{OD}} = \overrightarrow{\text{OC}} - \left( \overrightarrow{\text{OC}} + \overrightarrow{\text{CD}} \right) =\]

\[= - \overrightarrow{\text{CD}} = \overrightarrow{\text{CD}}.\]

\[\overrightarrow{\text{AB}} = \overrightarrow{\text{CD}}\ \]

\[(так\ как\ AB \parallel CD\ \ и\ \ AB = CD):\]

\[\overrightarrow{\text{OB}} - \overrightarrow{\text{OA}} = \overrightarrow{\text{OC}} - \overrightarrow{\text{OD}}.\]

\[Что\ и\ требовалось\ доказать.\]

\[\textbf{б)}\ \overrightarrow{\text{OB}} + \overrightarrow{\text{BC}} = \overrightarrow{\text{OC}}\ \]

\[(по\ правилу\ треугольника):\]

\[\overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{OB}} - \left( \overrightarrow{\text{OB}} + \overrightarrow{\text{BC}} \right) =\]

\[= - \overrightarrow{\text{BC}} = \overrightarrow{\text{CB}}.\]

\[\overrightarrow{\text{CB}} = \overrightarrow{\text{DA}}\ \]

\[(так\ как\ AB \parallel CD\ \ и\ \ AB = CD):\]

\[\overrightarrow{\text{OB}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{DA}}.\]

\[Что\ и\ требовалось\ доказать.\]