\[\boxed{\mathbf{1007.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[M \in AC;AM = MC;\]
\[N \in BD;BN = ND.\]
\[\mathbf{Доказать:}\]
\[MN = \frac{1}{2}(AD - BC).\]
\[\mathbf{Доказательство.}\]
\[1)\ По\ правилу\ многоугольника:\]
\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{DN}};\]
\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MC}} + \overrightarrow{\text{CB}} + \overrightarrow{\text{BN}}.\]
\[3)\ N\ и\ M - середины\ \text{BD\ }и\ AC:\]
\[\overrightarrow{\text{MA}} + \overrightarrow{\text{MC}} = \overrightarrow{0};\ \ \ \]
\[\overrightarrow{\text{DN}} + \overrightarrow{\text{BN}} = \overrightarrow{0}.\]
\[4)\ 2\overrightarrow{\text{MN}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}}\]
\[2\overrightarrow{\text{MN}} = \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}}\]
\[\overrightarrow{\text{MN}} = \frac{1}{2}\left( \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}} \right).\]
\[5)\ \frac{\overrightarrow{\text{AD}} \uparrow \uparrow \overrightarrow{\text{BC}}}{\overrightarrow{\text{MN}} \uparrow \uparrow \overrightarrow{\text{AD}}}\]
\[\left| \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}} \right| = AD - BC\]
\[MN = \frac{1}{2}(AD - BC).\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{1007.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ \overrightarrow{x}\ \left\{ - 3;\frac{1}{5} \right\}\ :\ \]
\[\overrightarrow{x} = - 3\overrightarrow{i} + \frac{1}{5}\overrightarrow{j}\text{\ .}\]
\[\textbf{б)}\ \overrightarrow{y}\ \left\{ - 2;\ - 3 \right\}\ :\ \]
\[\overrightarrow{y} = - 2\overrightarrow{i} - 3\overrightarrow{j}.\]
\[\textbf{в)}\ \overrightarrow{z}\ \left\{ - 1;0 \right\}\ :\ \]
\[\overrightarrow{z} = - \overrightarrow{i}.\]
\[\textbf{г)}\ \overrightarrow{u}\ \left\{ 0;3 \right\}\ :\ \]
\[\overrightarrow{u} = 3\overrightarrow{j}.\]
\[\textbf{д)}\ \overrightarrow{v}\ \left\{ 0;1 \right\}\ :\ \]
\[\overrightarrow{v} = \overrightarrow{j}.\]