ГДЗ по геометрии 11 класс. Атанасян ФГОС 622

Авторы:
Год:2023
Тип:учебник

622

\[\boxed{\mathbf{622.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - четырехугольник;\ \ \]

\[E,F,K,N - середины\ сторон\ \]

\[AB,BC,CD,DA;\ \ \]

\[M = EK \cap FN;\ \ \ \]

\[O - произвольная\ точка\ \]

\[пространства.\]

\[Доказать:\ \ \]

\[\overrightarrow{\text{OM}} = \frac{1}{4}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{OD}} \right).\]

\[Доказательство.\]

\[1)\ \overrightarrow{\text{OE}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BE}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AE}}:\]

\[2\overrightarrow{\text{OE}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BE}} + \overrightarrow{\text{OA}} + \overrightarrow{\text{AE}};\]

\[\overrightarrow{\text{BE}} = - \overrightarrow{\text{AE}}.\]

\[Отсюда:\]

\[2\overrightarrow{\text{OE}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}}\]

\[\overrightarrow{\text{OE}} = \frac{1}{2}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} \right).\]

\[2)\ Аналогично:\ \ \]

\[\overrightarrow{\text{OF}} = \frac{1}{2}\left( \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} \right);\ \]

\[\overrightarrow{\text{OK}} = \frac{1}{2}\left( \overrightarrow{\text{OC}} + \overrightarrow{\text{OD}} \right);\]

\[\overrightarrow{\text{ON}} = \frac{1}{2}\left( \overrightarrow{\text{OD}} + \overrightarrow{\text{OA}} \right)э\]

\[3)\ \overrightarrow{\text{OM}} = \overrightarrow{\text{OF}} + \overrightarrow{\text{FM}} =\]

\[= \overrightarrow{\text{ON}} + \overrightarrow{\text{NM}} = \overrightarrow{\text{OE}} + \overrightarrow{\text{EM}} =\]

\[= \overrightarrow{\text{OK}} + \overrightarrow{\text{KM}};\]

\[EF = NK = \frac{1}{2}\text{AC\ }\]

\[(как\ средние\ линии\ \mathrm{\Delta}ABC\ и\ \mathrm{\Delta}ADC).\]

\[Аналогично:\]

\[EN = FK.\]

\[EFKN - ромб;\]

\[\overrightarrow{\text{FM}} = - \overrightarrow{\text{NM}};\]

\[\overrightarrow{\text{EM}} = - \overrightarrow{\text{KM}}.\]

\[Следовательно:\]

\[4\overrightarrow{\text{OM}} = \overrightarrow{\text{OF}} + \overrightarrow{\text{OE}} + \overrightarrow{\text{OK}} + \overrightarrow{\text{ON}}\ \]

\[\overrightarrow{\text{OM}} =\]

\[= \frac{1}{4}\left( \overrightarrow{\text{OF}} + \overrightarrow{\text{OE}} + \overrightarrow{\text{OK}} + \overrightarrow{\text{ON}} \right) =\]

\[= \frac{1}{4}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{OD}} \right).\]

\[Что\ и\ требовалось\ доказать.\]

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