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МатематикаРешебник по геометрии 11 класс. Атанасян ФГОС 622
622
\[\boxed{\mathbf{622.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\ \ \]
\[ABCD - четырехугольник;\ \ \]
\[E,F,K,N - середины\ сторон\ \]
\[AB,BC,CD,DA;\ \ \]
\[M = EK \cap FN;\ \ \ \]
\[O - произвольная\ точка\ \]
\[пространства.\]
\[Доказать:\ \ \]
\[\overrightarrow{\text{OM}} = \frac{1}{4}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{OD}} \right).\]
\[Доказательство.\]
\[1)\ \overrightarrow{\text{OE}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BE}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AE}}:\]
\[2\overrightarrow{\text{OE}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BE}} + \overrightarrow{\text{OA}} + \overrightarrow{\text{AE}};\]
\[\overrightarrow{\text{BE}} = - \overrightarrow{\text{AE}}.\]
\[Отсюда:\]
\[2\overrightarrow{\text{OE}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}}\]
\[\overrightarrow{\text{OE}} = \frac{1}{2}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} \right).\]
\[2)\ Аналогично:\ \ \]
\[\overrightarrow{\text{OF}} = \frac{1}{2}\left( \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} \right);\ \]
\[\overrightarrow{\text{OK}} = \frac{1}{2}\left( \overrightarrow{\text{OC}} + \overrightarrow{\text{OD}} \right);\]
\[\overrightarrow{\text{ON}} = \frac{1}{2}\left( \overrightarrow{\text{OD}} + \overrightarrow{\text{OA}} \right)э\]
\[3)\ \overrightarrow{\text{OM}} = \overrightarrow{\text{OF}} + \overrightarrow{\text{FM}} =\]
\[= \overrightarrow{\text{ON}} + \overrightarrow{\text{NM}} = \overrightarrow{\text{OE}} + \overrightarrow{\text{EM}} =\]
\[= \overrightarrow{\text{OK}} + \overrightarrow{\text{KM}};\]
\[EF = NK = \frac{1}{2}\text{AC\ }\]
\[(как\ средние\ линии\ \mathrm{\Delta}ABC\ и\ \mathrm{\Delta}ADC).\]
\[Аналогично:\]
\[EN = FK.\]
\[EFKN - ромб;\]
\[\overrightarrow{\text{FM}} = - \overrightarrow{\text{NM}};\]
\[\overrightarrow{\text{EM}} = - \overrightarrow{\text{KM}}.\]
\[Следовательно:\]
\[4\overrightarrow{\text{OM}} = \overrightarrow{\text{OF}} + \overrightarrow{\text{OE}} + \overrightarrow{\text{OK}} + \overrightarrow{\text{ON}}\ \]
\[\overrightarrow{\text{OM}} =\]
\[= \frac{1}{4}\left( \overrightarrow{\text{OF}} + \overrightarrow{\text{OE}} + \overrightarrow{\text{OK}} + \overrightarrow{\text{ON}} \right) =\]
\[= \frac{1}{4}\left( \overrightarrow{\text{OA}} + \overrightarrow{\text{OB}} + \overrightarrow{\text{OC}} + \overrightarrow{\text{OD}} \right).\]
\[Что\ и\ требовалось\ доказать.\]
