\[\boxed{\mathbf{785.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[AM = MC;\ \ \]
\[BN = ND.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{MN}} = \frac{1}{2} \bullet \left( \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}} \right).\]
\[\mathbf{Доказательство.}\]
\[1)\ По\ правилу\ \]
\[многоугольников:\]
\[\ \overrightarrow{\text{MN}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{DN}}\]
\[+\]
\[\]
\[Отсюда:\]
\[\overrightarrow{\text{MA}} + \overrightarrow{\text{MC}} = \overrightarrow{0}\ \ \ и\ \ \ \overrightarrow{\text{DN}} + \overrightarrow{\text{BN}} = \overrightarrow{0};\]
\[так\ как:\]
\[AM = MC\ \ \ и\ \ \overrightarrow{\text{MA}} \nearrow \swarrow \overrightarrow{\text{MC}};\]
\[BN = DN\ \ \ и\ \ \overrightarrow{\text{DN}} \nearrow \swarrow \overrightarrow{\text{BN}}.\]
\[Следовательно:\]
\[2\overrightarrow{\text{MN}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}};\]
\[\overrightarrow{\text{MN}} = \frac{1}{2} \bullet \left( \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}} \right).\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]