\[\boxed{\mathbf{978.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{а})\ y = 3;\]
\[\textbf{б)}\ x = - 2;\]
\[\textbf{в)}\ y = - 4;\]
\[\textbf{г)}\ x = 7.\]
\[\boxed{\mathbf{978}\mathbf{.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - трапеция;\]
\[AC;BD - диагонали;\]
\[AM = MC;\]
\[BN = ND.\]
\[\mathbf{Доказать:}\]
\[MN = \frac{1}{2} \bullet (AD - BC).\]
\[\mathbf{Доказательство.}\]
\[По\ рисунку:\ \ \]
\[MN \parallel AD;\ MN \parallel BC.\]
\[1)\ По\ правилу\ \]
\[многоугольника:\]
\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{DN}};\]
\[\overrightarrow{\text{MN}} = \overrightarrow{\text{MC}} + \overrightarrow{\text{CB}} + \overrightarrow{\text{BN}}\text{.\ }\]
\[2)\ Складываем\ равенства:\]
\[Так\ как\ \ \overrightarrow{\text{MA}} + \overrightarrow{\text{MC}} = \overrightarrow{0};\ \]
\[\ \overrightarrow{\text{DN}} + \overrightarrow{\text{BN}} = \overrightarrow{0}:\]
\[\overrightarrow{\text{MN}} = \frac{\overrightarrow{\text{AD}} - \overrightarrow{\text{BC}}}{2}.\]
\[3)\ \overrightarrow{\text{AD}} \nearrow \nearrow \overrightarrow{\text{BC}};\ \ \]
\[\ \overrightarrow{\text{MN}} \nearrow \nearrow \frac{\overrightarrow{\text{AD}} - \overrightarrow{\text{BC}}}{2}:\]
\[\overrightarrow{\text{MN}} \parallel \overrightarrow{\text{AD}} \parallel \overrightarrow{\text{BC}};\ \ \ \]
\[\left| \overrightarrow{\text{MN}} \right| = \frac{1}{2}\left| \overrightarrow{\text{AD}} - \overrightarrow{\text{BC}} \right| = \frac{\overrightarrow{\text{AD}} - \overrightarrow{\text{BC}}}{2}.\]
\[Что\ и\ требовалось\ доказать.\]