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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 578
578
\[\boxed{\mathbf{578.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[ABCDP - пирамида;\ \ \]
\[ABCD - трапеция;\]
\[MN - средняя\ линия\ ABCD;\ \ \]
\[MO = ON.\]
\[Доказать:\]
\[\overrightarrow{\text{PA}} + \overrightarrow{\text{PB}} + \overrightarrow{\text{PC}} + \overrightarrow{\text{PD}} = 4\overrightarrow{\text{PO}}.\]
\[Доказательство.\]
\[1)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PA}} + \overrightarrow{\text{AO}};\ \ \]
\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PB}} + \overrightarrow{\text{BO}};\ \ \]
\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PC}} + \overrightarrow{\text{CO}};\ \ \]
\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PD}} + \overrightarrow{\text{DO}}.\]
\[Отсюда:\]
\[2)\ По\ правилу\ треугольника:\]
\[\overrightarrow{\text{AO}} = \overrightarrow{\text{AM}} + \overrightarrow{\text{MO}};\ \ \]
\[\overrightarrow{\text{BO}} = \overrightarrow{\text{BN}} + \overrightarrow{\text{NO}};\]
\[\overrightarrow{\text{MO}} = \overrightarrow{\text{MD}} + \overrightarrow{\text{DO}} \rightarrow \ \overrightarrow{\text{DO}} =\]
\[= \overrightarrow{\text{MO}} - \overrightarrow{\text{MD}};\ \ \]
\[\overrightarrow{\text{NO}} = \overrightarrow{\text{NC}} + \overrightarrow{\text{CO}} \rightarrow \ \overrightarrow{\text{CO}} =\]
\[= \overrightarrow{\text{NO}} - \overrightarrow{\text{NC}}.\]
\[3)\ MO\ и\ \text{NO\ }лежат\ на\ одной\ \]
\[прямой\ и\ MO = NO:\]
\[\overrightarrow{\text{MO}} = - \overrightarrow{\text{NO}}.\]
\[Аналогично:\]
\[\overrightarrow{\text{AM}} = \overrightarrow{\text{MD}}\text{\ \ }и\ \ \overrightarrow{\text{BN}} = \overrightarrow{\text{NC}}.\]
\[Отсюда:\ \]
\[\overrightarrow{\text{AO}} + \overrightarrow{\text{BO}} + \overrightarrow{\text{CO}} + \overrightarrow{\text{DO}} =\]
\[4)\ Получили:\]
\[4\overrightarrow{\text{PO}} = \overrightarrow{\text{PA}} + \overrightarrow{\text{PB}} + \overrightarrow{\text{PC}} + \overrightarrow{\text{PD}}.\]
\[Что\ и\ требовалось\ доказать.\]
