Решебник по геометрии 10 класс Атанасян ФГОС 623

\[\boxed{\mathbf{623.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[ABCD - параллелограмм;\ \]

\[AC \cap BD = O;\]

\[M - произвольная\ точка\ \ \]

\[пространства.\]

\[Доказать:\ \ \]

\[MO < \frac{1}{4}(MA + MB + MC + MD).\]

\[Доказательство.\]

\[1)\ \ ABCD - параллелограмм:\ \]

\[BO = OC;\ \]

\[CO = OA.\]

\[Отсюда:\]

\[\overrightarrow{\text{BO}} = - \overrightarrow{\text{DO}};\ \ \]

\[\overrightarrow{\text{AO}} = - \overrightarrow{\text{CO}}.\]

\[2)\ \overrightarrow{\text{MO}} = \overrightarrow{\text{MC}} + \overrightarrow{\text{CO}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{AO}} =\]

\[= \overrightarrow{\text{MB}} + \overrightarrow{\text{BO}} = \overrightarrow{\text{MD}} + \overrightarrow{\text{DO}}:\]

\[4\overrightarrow{\text{MO}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{MB}} + \overrightarrow{\text{MC}} + \overrightarrow{\text{MD}}\ \]

\[\overrightarrow{\text{MO}} = \frac{1}{4}\left( \overrightarrow{\text{MA}} + \overrightarrow{\text{MB}} + \overrightarrow{\text{MC}} + \overrightarrow{\text{MD}} \right).\]

\[3)\ Векторы\ \frac{1}{4}\overrightarrow{\text{MA}};\ \frac{1}{4}\overrightarrow{\text{MB}};\ \frac{1}{4}\overrightarrow{\text{MC}};\ \]

\[\frac{1}{4}\overrightarrow{\text{MD}}\ попарно\ не\ \]

\[сонаправлены:по\ доказанному\ \]

\[в\ задаче\ 587\]

\[Что\ и\ требовалось\ доказать.\ \]