Решебник по геометрии 11 класс. Атанасян ФГОС 109

\[\boxed{\mathbf{109.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} -\]

\[параллелепипед;\]

\[AA_{1}C_{1}C \cap \cap BDD_{1}B_{1} = a.\]

\[Доказать:\]

\[a \parallel AA_{1};\]

\[a \cap BD_{1};\]

\[a \cap B_{1}D;\]

\[a \cap AC_{1};\]

\[a \cap A_{1}\text{C.}\]

\[Доказательство.\]

\[1)\ AA_{1}\ не\ принадлежит\ \text{BD}D_{1};\ \]

\[BB_{1} \in BDD_{1};\ AA_{1} \parallel BB_{1}:\]

\[AA \parallel BDD_{1}.\]

\[2)\ AA_{1} \in ACC_{1};\ \ AA_{1} \parallel BDD_{1};\ \ \]

\[\text{AC}C_{1} \cap BDD_{1} = a:\]

\[a \parallel AA_{1}.\]

\[3)\ ABCDA_{1}B_{1}C_{1}D_{1} -\]

\[параллелепипед:\]

\[AA_{1} \parallel BB_{1} \parallel CC_{1} \parallel DD_{1}.\]

\[4)\ a \parallel AA_{1}:\]

\[a \parallel BB_{1};\ \ a \parallel CC_{1};\ \ \ a \parallel DD_{1}.\]

\[5)\ BB_{1} \cap B_{1}D;\ \ BB_{1} \cap BD_{1};\ \]

\[a \parallel BB_{1}:\]

\[a \cap BD_{1}\ и\ a \cap B_{1}\text{D\ }\]

\[(по\ теореме\ 3).\]

\[6)\ AA_{1} \cap AC_{1};\ \ AA_{1} \cap A_{1}C;\ \ \]

\[a \parallel AA_{1}:\]

\[a \cap AC_{1};\ \ \ a \cap A_{1}\text{C.}\]

\[Что\ и\ требовалось\ доказать.\]