Решебник по геометрии 10 класс Атанасян ФГОС 718

\[\boxed{\mathbf{718.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \]

\[ABCD - квадрат;\]

\[K_{1} - пересечение\ диагоналей;\]

\[AB = BC = CD = DA = a.\]

\[Доказать:\]

\[\angle(AK;BD) = 90{^\circ}.\]

\[Доказательство.\]

\[1)\ \overrightarrow{\text{AD}}\left\{ a;0;0 \right\};\ \ \ \overrightarrow{\text{AB}}\left\{ 0;a;0 \right\};\ \ \]

\[\overrightarrow{\text{BD}}\left\{ a; - a;0 \right\}.\]

\[K_{1}\left( \frac{a}{2};\frac{a}{2};0 \right);K\left( \frac{a}{2};\frac{a}{2};b \right);\ \ \]

\[KK_{1} = b;\]

\[\overrightarrow{\text{AK}}\left\{ \frac{a}{2};\frac{a}{2};b \right\}.\]

\[2)\ В\ треугольнике\ BAD:\]

\[\angle A = 90{^\circ};\]

\[BD = \sqrt{a^{2} + a^{2}} = a\sqrt{2}.\]

\[3)\ В\ треугольнике\ AKK_{1}:\]

\[\angle K_{1} = 90{^\circ};\]

\[AK_{1} = \frac{1}{2}BD = \frac{a\sqrt{2}}{2};\]

\[AK = \sqrt{b^{2} + \left( \frac{a\sqrt{2}}{2} \right)^{2}} =\]

\[= \sqrt{\frac{4b^{2} + 2a^{2}}{4}} = \sqrt{b^{2} + \frac{a^{2}}{2}}.\]

\[4)\ \overrightarrow{\text{AK}} \cdot \overrightarrow{\text{BD}} =\]

\[= \left| \overrightarrow{\text{AK}} \right| \cdot \left| \overrightarrow{\text{BD}} \right| \cdot \cos{\angle\left( \overrightarrow{\text{AK}};\overrightarrow{\text{BD}} \right)};\]

\[\overrightarrow{\text{AK}} \cdot \overrightarrow{\text{BD}} =\]

\[= x_{1} \cdot x_{2} + y_{1} \cdot y_{2} + z_{1} \cdot z_{2} =\]

\[= \frac{a^{2}}{2} - \frac{a^{2}}{2} = 0\]

\[a\sqrt{2} \cdot \sqrt{b^{2} + \frac{a^{2}}{2}} \cdot \cos{\angle\left( \overrightarrow{\text{AK}};\overrightarrow{\text{BD}} \right)} =\]

\[= 0.\]

\[Векторы\ не\ нулевые,\ косинус\ \]

\[равен\ 0:\]

\[\angle\left( \overrightarrow{\text{AK}};\overrightarrow{\text{BD}} \right) = 90{^\circ}.\]

\[Что\ и\ требовалось\ доказать.\]

\[Параграф\ 3.\ Движения\]