Решебник по геометрии 10 класс Атанасян ФГОС 611

\[\boxed{\mathbf{611.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \ \]

\[отрезки\ AB\ и\ \text{CD\ }не\ лежат\ в\ \]

\[одной\ плоскости;\]

\[точки\ \text{M\ }и\ N - середины\ AB\ и\ \]

\[\text{CD.}\]

\[Доказать:\ \ \]

\[MN < \frac{1}{2}(AC + BD).\]

\[Доказательство.\]

\[1)\ Точка\ M - середина\ AB;\ \]

\[\overrightarrow{\text{AM}} = - \overrightarrow{\text{BM}}:\]

\[\overrightarrow{\text{NM}} = \frac{1}{2}\left( \overrightarrow{\text{NA}} + \overrightarrow{\text{NB}} \right).\]

\[2)\ \overrightarrow{\text{NA}} = \overrightarrow{\text{NC}} + \overrightarrow{\text{CA}};\ \ \ \]

\[\overrightarrow{\text{NB}} = \overrightarrow{\text{ND}} + \overrightarrow{\text{DB}}:\]

\[\overrightarrow{\text{NM}} = \frac{1}{2}\left( \overrightarrow{\text{NC}} + \overrightarrow{\text{CA}} + \overrightarrow{\text{ND}} + \overrightarrow{\text{DB}} \right).\]

\[\overrightarrow{\text{NC}} = - \overrightarrow{\text{ND}}:\]

\[\overrightarrow{\text{NM}} = \frac{1}{2}\left( \overrightarrow{\text{CA}} + \overrightarrow{\text{DB}} \right).\]

\[3)\ Векторы\ \overrightarrow{\text{CA}}\ и\ \overrightarrow{\text{DB}} - не\ \]

\[коллинеарны\ (так\ как\ \text{CD\ }и\ \text{AB\ }\]

\[не\ лежат\ в\ одной\ плоскости,\ то\ \]

\[точки\ A,B,C\ и\ D - не\ лежат\ в\ \]

\[одной\ плоскости):\]

\[\overrightarrow{\text{CA}} + \overrightarrow{\text{DB}} < \left| \overrightarrow{\text{CA}} \right| + \left| \overrightarrow{\text{DB}} \right|\ \]

\[4)\ Таким\ образом:\ \]

\[MN < \frac{1}{2}(AC + BD).\]

\[Что\ и\ требовалось\ доказать.\]