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МатематикаГДЗ по геометрии 8 класс Атанасян Задание 807
Задание 807
\[\boxed{\mathbf{807.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AA_{1};\ \ BB_{1};\ \ CC_{1} - медианы;\]
\[точка\ O - произвольная.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} =\]
\[= \overrightarrow{OA_{1}} + \overrightarrow{OB_{1}} + \overrightarrow{OC_{1}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ По\ правилу\ треугольника:\]
\[\overrightarrow{OA_{1}} = \overrightarrow{\text{OA}} + \overrightarrow{AA_{1}};\]
\[\overrightarrow{OB_{1}} = \overrightarrow{\text{OB}} + \overrightarrow{BB_{1}};\]
\[\overrightarrow{OC_{1}} = \overrightarrow{\text{OC}} + \overrightarrow{CC_{1}}.\]
\[2)\ Сложим\ полученные\ \]
\[равенства:\]
\[3)\ Докажем,\ что\ \]
\[\ \overrightarrow{AA_{1}} + \overrightarrow{BB_{1}} + \overrightarrow{CC_{1}} = \overrightarrow{0}\text{.\ }\]
\[По\ условию:\]
\[BA_{1} = A_{1}C;\ \ B_{1}C = AB_{1};\ \ \]
\[AC_{1} = C_{1}\text{B.}\]
\[Получаем:\]
\[\overrightarrow{BB_{1}} = \overrightarrow{\text{BA}} + \overrightarrow{AB_{1}} = - \overrightarrow{\text{AB}} + \frac{1}{2}\overrightarrow{\text{AC}}.\]
\[\overrightarrow{CC_{1}} = \overrightarrow{\text{CA}} + \overrightarrow{AC_{1}} = - \overrightarrow{\text{AC}} + \frac{1}{2}\overrightarrow{\text{AB}}.\]
\[\overrightarrow{AA_{1}} = \overrightarrow{\text{AB}} + \overrightarrow{BA_{1}} = \overrightarrow{\text{AB}} + \frac{1}{2}\overrightarrow{\text{BC}}.\]
\[\overrightarrow{\text{BC}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AC}} = - \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}}.\]
\[\overrightarrow{AA_{1}} = \overrightarrow{\text{AB}} + \frac{1}{2} \bullet \left( - \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} \right) =\]
\[= \frac{1}{2}\overrightarrow{\text{AB}} + \frac{1}{2}\overrightarrow{\text{AC}}.\]
\[Подставляем:\]
\[= \overrightarrow{\text{AB}} - \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} - \overrightarrow{\text{AC}} =\]
\[= \overrightarrow{\text{BB}} + \overrightarrow{\text{CC}} = \overrightarrow{0.}\]
\[4)\ \overrightarrow{OA_{1}} + \overrightarrow{OB_{1}} + \overrightarrow{OC_{1}} =\]
\[= \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} + \overrightarrow{0}\]
\[\overrightarrow{OA_{1}} + \overrightarrow{OB_{1}} + \overrightarrow{OC_{1}} =\]
\[= \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]
