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МатематикаГДЗ по геометрии 11 класс. Атанасян ФГОС 748
748
\[\boxed{\mathbf{748.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[Доказательство.\]
\[1)\ E_{2}E_{3} - средняя\ линия\ грани\ \]
\[ABD:\]
\[E_{2}E_{3} = \frac{\text{DB}}{2}.\]
\[E_{1}E_{4} - средняя\ линия\ грани\ \]
\[BCD:\]
\[E_{1}E_{4} = \frac{\text{DB}}{2}.\]
\[2)\ \overrightarrow{E_{4}O} = \overrightarrow{E_{4}E_{1}} + \overrightarrow{E_{1}O} =\]
\[= \frac{1}{2}\overrightarrow{\text{DB}} + \overrightarrow{E_{1}O};\]
\[\overrightarrow{OE_{1}} = \overrightarrow{OE_{2}} + \overrightarrow{E_{2}E_{3}} =\]
\[= \overrightarrow{OE_{2}} + \frac{1}{2}\overrightarrow{\text{DB}}.\]
\[По\ условию\ \overrightarrow{OE_{2}} = \overrightarrow{E_{1}O}:\]
\[\overrightarrow{E_{4}O} = \overrightarrow{OE_{3}};\]
\[O - середина\ отрезка\ E_{3}E_{4}.\]
\[3)\ \overrightarrow{\text{DO}} = \overrightarrow{DE_{2}} + \overrightarrow{E_{2}O} =\]
\[= \frac{1}{2}\overrightarrow{\text{DA}} + \frac{1}{2}\overrightarrow{E_{2}E_{1}}\]
\[\overrightarrow{E_{2}E_{1}} = \overrightarrow{E_{2}D} + \overrightarrow{\text{DC}} + \overrightarrow{CE_{1}};\]
\[\overrightarrow{E_{2}E_{1}} = \overrightarrow{E_{2}A} + \overrightarrow{\text{AB}} + \overrightarrow{BE_{1}}.\]
\[Складываем\ два\ последних\ \]
\[равенства,\ учитывая,\ что\ \]
\[\overrightarrow{E_{2}D} + \overrightarrow{E_{2}A} = 0;\]
\[\overrightarrow{E_{1}C} + \overrightarrow{E_{1}B} = 0:\]
\[2\overrightarrow{E_{2}E_{1}} =\]
\[= \overrightarrow{\text{DC}} + \overrightarrow{\text{AB}}\]
\[\overrightarrow{E_{2}E_{1}} = \frac{1}{2}\left( \overrightarrow{\text{DC}} + \overrightarrow{\text{AB}} \right).\]
\[4)\ \overrightarrow{\text{DO}} = \frac{\overrightarrow{\text{DA}}}{2} + \frac{\overrightarrow{E_{2}E_{1}}}{2} =\]
\[= \frac{\overrightarrow{\text{DA}}}{2} + \frac{\overrightarrow{\text{DC}} + \overrightarrow{\text{AB}}}{4} =\]
\[= \frac{\overrightarrow{\text{DA}}}{2} + \frac{\overrightarrow{\text{DC}}}{4} + \frac{\overrightarrow{\text{AB}}}{4} =\]
\[= \frac{\overrightarrow{\text{DA}}}{2} + \frac{\overrightarrow{\text{DC}}}{4} + \frac{\overrightarrow{\text{DB}} - \overrightarrow{\text{DA}}}{4} =\]
\[= \frac{\overrightarrow{\text{DA}} + \overrightarrow{\text{DB}} + \overrightarrow{\text{DC}}}{4};\]
\[\overrightarrow{\text{DM}} = \overrightarrow{\text{DA}} + \overrightarrow{\text{AM}} = \overrightarrow{\text{DA}} + \frac{2}{3}\overrightarrow{AE_{1}} =\]
\[= \overrightarrow{\text{DA}} + \frac{1}{3}\left( \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} \right) =\]
\[= \overrightarrow{\text{DA}} + \frac{1}{3}\left( \overrightarrow{\text{DB}} + \overrightarrow{\text{DC}} - 2\overrightarrow{\text{DA}} \right) =\]
\[= \frac{1}{3}\overrightarrow{\text{DB}} + \frac{1}{3}\overrightarrow{\text{DC}} + \frac{1}{3}\overrightarrow{\text{DA}} =\]
\[= \frac{1}{3}\left( \overrightarrow{\text{DB}} + \overrightarrow{\text{DC}} + \overrightarrow{\text{DA}} \right).\]
\[Отсюда:\ \]
\[\overrightarrow{\text{DO}} = \frac{3}{4}\overrightarrow{\text{DM}};\]
\[\overrightarrow{\text{OM}} = \frac{1}{4}\overrightarrow{\text{DM}};\]
\[\frac{\overrightarrow{\text{DO}}}{\overrightarrow{\text{OM}}} = 3.\]
\[Следовательно:\]
\[точка\ O\ делит\ \text{DM\ }в\ отношении\ \]
\[3\ :1,\ считая\ от\ вершины.\]
\[Для\ других\ медиан\ \]
\[доказательство\ аналогичное.\]
\[Что\ и\ требовалось\ доказать.\]
