Решебник по геометрии 10 класс Атанасян ФГОС 598

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\[1)\ \overrightarrow{\text{CD}} = - \overrightarrow{\text{AB}}:\]

\[\overrightarrow{\text{CD}} = 0 \bullet \overrightarrow{\text{AD}} - 1 \bullet \overrightarrow{\text{AB}} + 0 \bullet \overrightarrow{AA_{1}}.\]

\[2)\ По\ правилу\ \]

\[параллелограмма:\]

\[\overrightarrow{AA_{1}} + \overrightarrow{\text{AD}} = \overrightarrow{AD_{1}}.\]

\[3)\ \overrightarrow{AD_{1}} + \overrightarrow{D_{1}B} = \overrightarrow{\text{AB}};\ \ \text{\ \ }\]

\[\overrightarrow{D_{1}B} = 2 \bullet \overrightarrow{D_{1}O}:\]

\[\overrightarrow{AD_{1}} + 2\overrightarrow{D_{1}O} = \overrightarrow{\text{AB}}.\]

\[Отсюда:\ \]

\[\overrightarrow{D_{1}O} = \frac{1}{2}\overrightarrow{\text{AB}} - \frac{1}{2}\overrightarrow{AD_{1}} =\]

\[= \frac{1}{2}\overrightarrow{\text{AB}} - \frac{1}{2}\left( \overrightarrow{AA_{1}} + \overrightarrow{\text{AD}} \right).\]