\[\boxed{\mathbf{772.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{XA}} + \overrightarrow{\text{XC}} = \overrightarrow{\text{XB}} + \overrightarrow{\text{XD}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ \overrightarrow{\text{XA}} - \overrightarrow{\text{XB}} = \overrightarrow{\text{BX}} + \overrightarrow{\text{XA}} = \overrightarrow{\text{BA}};\]
\[\overrightarrow{\text{XD}} - \overrightarrow{\text{XC}} = \overrightarrow{\text{CX}} + \overrightarrow{\text{XD}} =\]
\[= \overrightarrow{\text{CD}}\text{\ \ }\]
\[(по\ правилу\ треугольника).\]
\[2)\ ABCD - параллелограмм:\]
\[\overrightarrow{\text{BA}} \uparrow \uparrow \overrightarrow{\text{CD}}\text{\ \ }и\ \left| \overrightarrow{\text{BA}} \right| = \left| \overrightarrow{\text{CD}} \right|.\]
\[Следовательно:\ \]
\[\overrightarrow{\text{BA}} = \overrightarrow{\text{CD}}.\]
\[3)\ Получаем:\ \]
\[\overrightarrow{\text{XA}} - \overrightarrow{\text{XB}} = \overrightarrow{\text{XD}} - \overrightarrow{\text{XC}};\]
\[\overrightarrow{\text{XA}} + \overrightarrow{\text{XC}} = \overrightarrow{\text{XB}} + \overrightarrow{\text{XD}}.\]
\[Что\ и\ требовалось\ доказать.\]