ГДЗ по геометрии 8 класс Атанасян Задание 908

\[\boxed{\mathbf{908.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - четырехугольник;\]

\[M,N,P,Q - середины;\]

\[MP \cap NQ = O;\]

\[AE = EC;\]

\[BF = FD.\]

\[\mathbf{Доказать:}\]

\[O \in EF.\]

\[\mathbf{Доказательство.}\]

\[1)\ Определим\ вид\ \text{MNPQ.}\]

\[Пусть\ \overrightarrow{\text{AM}} = \overrightarrow{\text{MB}} = \overrightarrow{a},\]

\[\ \overrightarrow{\text{BN}} = \overrightarrow{\text{NC}} = \overrightarrow{b};\ \overrightarrow{\text{CP}} = \overrightarrow{\text{PD}} = \overrightarrow{c},\]

\[\ \overrightarrow{\text{DQ}} = \overrightarrow{\text{QA}} = \overrightarrow{d}:\]

\[\overrightarrow{\text{AB}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}}\]

\[2\overrightarrow{a} = - 2\overrightarrow{d} - 2\overrightarrow{c} - 2\overrightarrow{b}\]

\[\overrightarrow{a} + \overrightarrow{b} = - \left( \overrightarrow{c} + \overrightarrow{d} \right).\]

\[Отсюда:\ \]

\[\overrightarrow{\text{MN}} = - \overrightarrow{\text{QP}} = \overrightarrow{\text{PQ}};\]

\[MN \parallel QP\ и\ MN = QP;\ \]

\[MNPQ - параллелограмм.\]

\[2)\ Получаем:\ \]

\[\overrightarrow{\text{MN}} = \overrightarrow{\text{QP}} = \overrightarrow{m},\ \]

\[\overrightarrow{\text{NP}} = \overrightarrow{\text{MQ}} = \overrightarrow{n},\ \]

\[\overrightarrow{\text{MO}} = \overrightarrow{\text{OP}} = \overrightarrow{p},\]

\[\overrightarrow{\text{NO}} = \overrightarrow{\text{OQ}} = \overrightarrow{q}.\]

\[3)\ В\ треугольнике\ ABD:\]

\[MQ - средняя\ линия \Longrightarrow \ \]

\[\overrightarrow{\text{BD}} = 2\overrightarrow{\text{MQ}} = 2\overrightarrow{n};\]

\[\overrightarrow{\text{BF}} = \overrightarrow{\text{FD}} = \overrightarrow{n}.\]

\[4)\ В\ треугольнике\ ABC:\]

\[MN - средняя\ линия \Longrightarrow\]

\[\overrightarrow{\text{AC}} = 2\overrightarrow{\text{MN}} = 2\overrightarrow{m};\]

\[\overrightarrow{\text{AE}} = \overrightarrow{\text{EC}} = \overrightarrow{m}.\]

\[5)\ \overrightarrow{\text{EO}} = \overrightarrow{\text{AO}} - \overrightarrow{\text{AE}} =\]

\[= \left( \overrightarrow{\text{AM}} + \overrightarrow{\text{MO}} \right) - \overrightarrow{\text{AE}} =\]

\[= \overrightarrow{a} + \overrightarrow{p} - \overrightarrow{m};\]

\[\overrightarrow{\text{OF}} = \overrightarrow{\text{BF}} - \overrightarrow{\text{BO}} =\]

\[= \overrightarrow{\text{BF}} - \left( \overrightarrow{\text{BN}} + \overrightarrow{\text{NO}} \right) =\]

\[= \overrightarrow{n} - \left( \overrightarrow{n} + \overrightarrow{q} \right).\]

\[\overrightarrow{\text{EO}} - \overrightarrow{\text{OF}} =\]

\[= \overrightarrow{a} + \overrightarrow{p} - \overrightarrow{m} - \left( \overrightarrow{n} - \left( \overrightarrow{b} + \overrightarrow{q} \right) \right) =\]

\[= \overrightarrow{a} + \overrightarrow{b} - \left( \overrightarrow{m} + \overrightarrow{n} \right) + \overrightarrow{p} + \overrightarrow{q} =\]

\[= \overrightarrow{m} - \overrightarrow{m} - \overrightarrow{n} + \overrightarrow{n} = \overrightarrow{0}.\]

\[6)\ Таким\ образом:\ \]

\[\overrightarrow{\text{EO}} = \overrightarrow{\text{OF}} \Longrightarrow \ O \in EF.\]

\[Что\ и\ требовалось\ доказать.\]