ГДЗ по геометрии 9 класс Атанасян Задание 1007

\[\boxed{\mathbf{1007.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - трапеция;\]

\[M \in AC;AM = MC;\]

\[N \in BD;BN = ND.\]

\[\mathbf{Доказать:}\]

\[MN = \frac{1}{2}(AD - BC).\]

\[\mathbf{Доказательство.}\]

\[1)\ По\ правилу\ многоугольника:\]

\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{DN}};\]

\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MC}} + \overrightarrow{\text{CB}} + \overrightarrow{\text{BN}}.\]

\[3)\ N\ и\ M - середины\ \text{BD\ }и\ AC:\]

\[\overrightarrow{\text{MA}} + \overrightarrow{\text{MC}} = \overrightarrow{0};\ \ \ \]

\[\overrightarrow{\text{DN}} + \overrightarrow{\text{BN}} = \overrightarrow{0}.\]

\[4)\ 2\overrightarrow{\text{MN}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}}\]

\[2\overrightarrow{\text{MN}} = \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}}\]

\[\overrightarrow{\text{MN}} = \frac{1}{2}\left( \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}} \right).\]

\[5)\ \frac{\overrightarrow{\text{AD}} \uparrow \uparrow \overrightarrow{\text{BC}}}{\overrightarrow{\text{MN}} \uparrow \uparrow \overrightarrow{\text{AD}}}\]

\[\left| \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}} \right| = AD - BC\]

\[MN = \frac{1}{2}(AD - BC).\]

\[Что\ и\ требовалось\ доказать.\]