ГДЗ по геометрии 8 класс Атанасян Задание 855

\[\boxed{\mathbf{855.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\ \]

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

\[CD\bot AB;\]

\[DE\bot AC;\]

\[DF\bot BC.\]

\[\mathbf{Доказать:}\]

\[\textbf{а)}\ CD^{3} = AB \cdot AE \cdot BF\]

\[\textbf{б)}AE^{2} + BF^{2} + 3CD^{2} = AB^{2}\]

\[\textbf{в)}\ \sqrt[3]{AE^{2}} + \sqrt[3]{BF^{2}} = \sqrt[3]{AB^{2}}\]

\[\mathbf{Доказательство.}\]

\[Подобные\ треугольники,\ \]

\[полученные\ в\ результате\ \]

\[построения:\ \]

\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}ADE\sim\mathrm{\Delta}ACD\sim\mathrm{\Delta}DCB\sim\mathrm{\Delta}DBF.\]

\[\textbf{а)}\ \frac{\text{AD}}{\text{CD}} = \frac{\text{CD}}{\text{BD}} \Longrightarrow CD^{2} = AD \cdot BD;\]

\[\frac{\text{AB}}{\text{BC}} = \frac{\text{AC}}{\text{CD}} \Longrightarrow CD = \frac{AC \cdot BC}{\text{AB}};\]

\[\frac{\text{AD}}{\text{AC}} = \frac{\text{DE}}{\text{CD}} \Longrightarrow CD = \frac{AC \cdot DE}{\text{AD}};\]

\[CD =\]

\[= AD \cdot BD \cdot \frac{AC \cdot BC}{\text{AB}} \cdot \frac{AC \cdot DE}{\text{AD}} =\]

\[= AC^{2} \cdot BD \cdot BC \cdot \frac{\text{DE}}{\text{AB}}.\]

\[\frac{\text{AC}}{\text{AD}} = \frac{\text{AB}}{\text{AC}} \Longrightarrow AC^{2} = AB \cdot AD;\]

\[CD^{4} = AD \cdot BD \cdot DC \cdot DE.\]

\[\frac{\text{AD}}{\text{AB}} = \frac{\text{DE}}{\text{BC}} \Longrightarrow AD \cdot BC = AB \cdot DE;\]

\[\frac{\text{DE}}{\text{BF}} = \frac{\text{AE}}{\text{DF}} \Longrightarrow DE = \frac{AE \cdot BF}{\text{DF}};\]

\[CD^{4} = AB \cdot BD \cdot DE^{2} =\]

\[= AB \cdot BD \cdot DE \cdot \frac{AE \cdot BF}{\text{DF}} =\]

\[= AB \cdot AE \cdot BF \cdot \frac{BD \cdot DE}{\text{DF}} =\]

\[= AB \cdot AE \cdot BF \cdot \frac{AC \cdot DE}{\text{AD}} =\]

\[= AB \cdot AE \cdot BF \cdot CD.\]

\[CD^{3} = AB \cdot AE \cdot BF.\]

\[\textbf{б)}\ CD^{2} = AD \cdot BD + теорема\ \]

\[Пифагора:\ \]

\[AB^{2} = (AD + BD)^{2} =\]

\[= AD^{2} + BD^{2} + 2AD \cdot BD =\]

\[= AD^{2} + BD^{2} + 2CD^{2} =\]

\[= AE^{2} + BF^{2} + 3CD^{2}.\]

\[\textbf{в)}\ \frac{\text{AD}}{\text{AE}} = \frac{\text{AB}}{\text{AC}} \Longrightarrow AD = \frac{AB \cdot AE}{\text{AC}}.\]

\[\frac{\text{AC}}{\text{AD}} = \frac{\text{AD}}{\text{AE}} \Longrightarrow AD^{2} = AC \cdot AE.\]

\[AD^{3} = \frac{AB \cdot AE}{\text{AC}} \cdot AC \cdot AE =\]

\[= AB \cdot AE^{2}.\]

\[\frac{\text{AB}}{\text{BD}} = \frac{\text{BC}}{\text{BF}} \Longrightarrow \ BD = \frac{AB \cdot BF}{\text{BC}}.\]

\[\frac{\text{BC}}{\text{BD}} = \frac{\text{BD}}{\text{BF}} \Longrightarrow BD^{2} = BC \cdot BF.\]

\[BD^{3} = \frac{AB \cdot BF}{\text{BC}} \cdot BC \cdot BF =\]

\[= AB \cdot BF^{2}.\]

\[AB = AD + BD =\]

\[= \sqrt[3]{AB \cdot AE^{2}} + \sqrt[3]{AB \cdot BF^{2}}\ \ \ |\ :\sqrt[3]{\text{AB}}\]

\[\sqrt[3]{AB^{2}} = \sqrt[3]{AE^{2}} + \sqrt[3]{BF^{2}}.\]

\[Что\ и\ требовалось\ доказать.\]