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МатематикаГДЗ по геометрии 7 класс Атанасян Задание 791
Задание 791
\[\boxed{\mathbf{791.}\mathbf{ОК}\mathbf{\ }\mathbf{ГДЗ}\mathbf{-}\mathbf{домашка}\mathbf{\ }\mathbf{на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[\text{BN} = \text{NC};\text{CF} = \text{FD};\ \]
\[BE = EA;AM = MD;\]
\[EF \cap NM = 0.\]
\[\mathbf{Доказать:}\]
\[NO = OM;\ \ \]
\[EO = OF.\]
\[\mathbf{Доказательство.}\]
\[1) - \overrightarrow{\text{EA}} + \overrightarrow{\text{AM}} + \overrightarrow{\text{MD}} + \overrightarrow{\text{DF}} =\]
\[= \overrightarrow{\text{EO}} + \overrightarrow{\text{OF}}\]
\[- \overrightarrow{\text{EA}} + \overrightarrow{\text{AM}} + \overrightarrow{\text{MO}} = \overrightarrow{\text{EO}}\]
\[- \overrightarrow{\text{EA}} + 2\overrightarrow{\text{AM}} + \overrightarrow{\text{DF}} = \overrightarrow{\text{EO}} + \overrightarrow{\text{OF}}\]
\[- \overrightarrow{\text{EA}} + \overrightarrow{\text{AM}} + \overrightarrow{\text{EO}} = \overrightarrow{\text{EO}}\text{.\ }\]
\[2) - \overrightarrow{\text{MA}} + \overrightarrow{\text{AE}} + \overrightarrow{\text{EB}} + \overrightarrow{\text{BN}} =\]
\[= \overrightarrow{\text{MO}} + \overrightarrow{\text{ON}}\]
\[- \overrightarrow{\text{MA}} + \overrightarrow{\text{AE}} + \overrightarrow{\text{EO}} = \overrightarrow{\text{MO}}\]
\[- \overrightarrow{\text{MA}} + 2\overrightarrow{\text{AE}} + \overrightarrow{\text{BN}} = \overrightarrow{\text{MO}} + \overrightarrow{\text{ON}}\]
\[- \overrightarrow{\text{MA}} + \overrightarrow{\text{AE}} + \overrightarrow{\text{EO}} = \overrightarrow{\text{MO}}.\]
\[3)\ Выразим\ \overrightarrow{\text{EO}}\text{\ \ }и\ \ \ \overrightarrow{\text{MO}};\]
\[подставим:\]
\[- \overrightarrow{\text{EA}} + 2\overrightarrow{\text{AM}} + \overrightarrow{\text{DF}} =\]
\[= - \overrightarrow{\text{EA}} + \overrightarrow{\text{AM}} + \overrightarrow{\text{MO}} + \overrightarrow{\text{OF}}\]
\[- \overrightarrow{\text{MA}} + 2\overrightarrow{\text{AE}} + \overrightarrow{\text{BN}} =\]
\[= - \overrightarrow{\text{MA}} + \overrightarrow{\text{AE}} + \overrightarrow{\text{EO}} + \overrightarrow{\text{ON}}.\]
\[Получим:\]
\[\overrightarrow{\text{AM}} + \overrightarrow{\text{DF}} = \overrightarrow{\text{MO}} + \overrightarrow{\text{OF}}\]
\[\overrightarrow{\text{AE}} + \overrightarrow{\text{BN}} = \overrightarrow{\text{EO}} + \overrightarrow{\text{ON}}.\]
\[4)\ Запишем\ выражения:\]
\[2\overrightarrow{\text{AM}} + 2\overrightarrow{\text{DF}} = \overrightarrow{\text{AC}}\]
\[2\overrightarrow{\text{AE}} + 2\overrightarrow{\text{BN}} = \overrightarrow{\text{AC}}\]
\[Получим\ равенство:\]
\[\overrightarrow{\text{AM}} + \overrightarrow{\text{DF}} = \overrightarrow{\text{AE}} + \overrightarrow{\text{BN}}.\]
\[Следовательно:\]
\[\overrightarrow{\text{MO}} + \overrightarrow{\text{OF}} = \overrightarrow{\text{EO}} + \overrightarrow{\text{ON}}\]
\[\overrightarrow{\text{MO}} - \overrightarrow{\text{ON}} = \overrightarrow{\text{EO}} - \overrightarrow{\text{OF}}.\]
\[5)\ Так\ как\ \ \overrightarrow{\text{MO}} \nearrow \nearrow \overrightarrow{\text{ON}};\ \]
\[\overrightarrow{\text{EO}} \nearrow \nearrow \overrightarrow{\text{OF}};\ EO;OF\ \ и\ \ MO;ON;\ \ \ \]
\[то\ не\ \ коллинеарные:\]
\[NO = OM;\ \ EO = OF.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]
