Решебник по геометрии 11 класс. Атанасян ФГОС 578

Авторы:
Год:2023
Тип:учебник

578

\[\boxed{\mathbf{578.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[ABCDP - пирамида;\ \ \]

\[ABCD - трапеция;\]

\[MN - средняя\ линия\ ABCD;\ \ \]

\[MO = ON.\]

\[Доказать:\]

\[\overrightarrow{\text{PA}} + \overrightarrow{\text{PB}} + \overrightarrow{\text{PC}} + \overrightarrow{\text{PD}} = 4\overrightarrow{\text{PO}}.\]

\[Доказательство.\]

\[1)\ По\ правилу\ треугольника:\]

\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PA}} + \overrightarrow{\text{AO}};\ \ \]

\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PB}} + \overrightarrow{\text{BO}};\ \ \]

\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PC}} + \overrightarrow{\text{CO}};\ \ \]

\[\overrightarrow{\text{PO}} = \overrightarrow{\text{PD}} + \overrightarrow{\text{DO}}.\]

\[Отсюда:\]

\[2)\ По\ правилу\ треугольника:\]

\[\overrightarrow{\text{AO}} = \overrightarrow{\text{AM}} + \overrightarrow{\text{MO}};\ \ \]

\[\overrightarrow{\text{BO}} = \overrightarrow{\text{BN}} + \overrightarrow{\text{NO}};\]

\[\overrightarrow{\text{MO}} = \overrightarrow{\text{MD}} + \overrightarrow{\text{DO}} \rightarrow \ \overrightarrow{\text{DO}} =\]

\[= \overrightarrow{\text{MO}} - \overrightarrow{\text{MD}};\ \ \]

\[\overrightarrow{\text{NO}} = \overrightarrow{\text{NC}} + \overrightarrow{\text{CO}} \rightarrow \ \overrightarrow{\text{CO}} =\]

\[= \overrightarrow{\text{NO}} - \overrightarrow{\text{NC}}.\]

\[3)\ MO\ и\ \text{NO\ }лежат\ на\ одной\ \]

\[прямой\ и\ MO = NO:\]

\[\overrightarrow{\text{MO}} = - \overrightarrow{\text{NO}}.\]

\[Аналогично:\]

\[\overrightarrow{\text{AM}} = \overrightarrow{\text{MD}}\text{\ \ }и\ \ \overrightarrow{\text{BN}} = \overrightarrow{\text{NC}}.\]

\[Отсюда:\ \]

\[\overrightarrow{\text{AO}} + \overrightarrow{\text{BO}} + \overrightarrow{\text{CO}} + \overrightarrow{\text{DO}} =\]

\[4)\ Получили:\]

\[4\overrightarrow{\text{PO}} = \overrightarrow{\text{PA}} + \overrightarrow{\text{PB}} + \overrightarrow{\text{PC}} + \overrightarrow{\text{PD}}.\]

\[Что\ и\ требовалось\ доказать.\]

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