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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 868
Задание 868
\[\boxed{\mathbf{868.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[N \in CD;\]
\[M = AN \cap BD;\]
\[P = AN \cap BC.\]
\[\mathbf{Доказать:}\]
\[AM = \sqrt{MN \bullet MP}.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ AB = CD = a;\ \ \]
\[AD = BC = b.\]
\[2)\ \mathrm{\Delta}BAM\sim\mathrm{\Delta}DNM\ \]
\[(по\ двум\ углам):\]
\[\frac{\text{AM}}{\text{MN}} = \frac{\text{AB}}{\text{ND}}\ \]
\[\frac{\text{AM}}{\text{MN}} = \frac{a}{a + NC}\]
\[\frac{\text{NC}}{a} + 1 = \frac{\text{MN}}{\text{AM}}.\]
\[3)\ \mathrm{\Delta}BMP\sim\mathrm{\Delta}DMA\ \]
\[(по\ двум\ углам):\]
\[\frac{\text{BP}}{\text{AD}} = \frac{\text{MP}}{\text{AM}};\ \ \]
\[MP = \frac{\text{BP}}{b}AM = \frac{b - PC}{b}AM =\]
\[= \left( 1 - \frac{\text{PC}}{b} \right)\text{AM.}\]
\[4)\ PC \parallel AD\ \]
\[(по\ теореме\ Фалеса):\]
\[\frac{\text{AP}}{\text{NP}} = \frac{\text{CD}}{\text{NC}} = \frac{a}{\text{NC}};\ \ \ \]
\[\frac{\text{AP}}{\text{NP}} = \frac{a}{\text{NC}};\ \ \]
\[\ \frac{\text{NC}}{a} = \frac{\text{NP}}{\text{AP}};\]
\[\frac{\text{PC}}{\text{AD}} = \frac{\text{NP}}{\text{AN}};\ \ \ \]
\[\frac{\text{PC}}{b} = \frac{\text{NP}}{\text{AN}}.\]
\[5)\ MP = \left( 1 - \frac{\text{PC}}{b} \right)AM =\]
\[= \left( 1 - \frac{\text{NP}}{\text{AN}} \right)AM =\]
\[= \frac{AN - NP}{\text{AN}}AM = \frac{\text{AP}}{\text{AN}}\text{AM.}\]
\[\frac{\text{NP}}{\text{AP}} + 1 = \frac{\text{MN}}{\text{AM}};\ \ \ \]
\[\frac{NP + AP}{\text{AP}} = \frac{\text{MN}}{\text{AM}};\ \ \ \]
\[\frac{\text{AN}}{\text{AP}} = \frac{\text{MN}}{\text{AM}}\]
\[\frac{\text{MN}}{\text{AM}} = \frac{\text{AN}}{\text{AP}} = \frac{\text{AM}}{\text{MP}}\]
\[AM^{2} = MN \bullet MP\]
\[AM = \sqrt{MN \bullet MP}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]
\[\boxed{\mathbf{868.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\ \]
\[\angle C = 90{^\circ};\]
\[CD\bot AB;\]
\[DE\bot AC;\]
\[DF\bot BC.\]
\[\mathbf{Доказать:}\]
\[\textbf{а)}\ CD^{3} = AB \cdot AE \cdot BF\]
\[\textbf{б)}AE^{2} + BF^{2} + 3CD^{2} = AB^{2}\]
\[\textbf{в)}\ \sqrt[3]{AE^{2}} + \sqrt[3]{BF^{2}} = \sqrt[3]{AB^{2}}\]
\[\mathbf{Доказательство.}\]
\[Подобные\ треугольники,\ \]
\[полученные\ в\ результате\ \]
\[построения:\ \]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}ADE\sim\mathrm{\Delta}ACD\sim\mathrm{\Delta}DCB\sim\mathrm{\Delta}DBF.\]
\[\textbf{а)}\ \frac{\text{AD}}{\text{CD}} = \frac{\text{CD}}{\text{BD}} \Longrightarrow CD^{2} = AD \cdot BD;\]
\[\frac{\text{AB}}{\text{BC}} = \frac{\text{AC}}{\text{CD}} \Longrightarrow CD = \frac{AC \cdot BC}{\text{AB}};\]
\[\frac{\text{AD}}{\text{AC}} = \frac{\text{DE}}{\text{CD}} \Longrightarrow CD = \frac{AC \cdot DE}{\text{AD}};\]
\[CD =\]
\[= AD \cdot BD \cdot \frac{AC \cdot BC}{\text{AB}} \cdot \frac{AC \cdot DE}{\text{AD}} =\]
\[= AC^{2} \cdot BD \cdot BC \cdot \frac{\text{DE}}{\text{AB}}.\]
\[\frac{\text{AC}}{\text{AD}} = \frac{\text{AB}}{\text{AC}} \Longrightarrow AC^{2} = AB \cdot AD;\]
\[CD^{4} = AD \cdot BD \cdot DC \cdot DE.\]
\[\frac{\text{AD}}{\text{AB}} = \frac{\text{DE}}{\text{BC}} \Longrightarrow AD \cdot BC = AB \cdot DE;\]
\[\frac{\text{DE}}{\text{BF}} = \frac{\text{AE}}{\text{DF}} \Longrightarrow DE = \frac{AE \cdot BF}{\text{DF}};\]
\[CD^{4} = AB \cdot BD \cdot DE^{2} =\]
\[= AB \cdot BD \cdot DE \cdot \frac{AE \cdot BF}{\text{DF}} =\]
\[= AB \cdot AE \cdot BF \cdot \frac{BD \cdot DE}{\text{DF}} =\]
\[= AB \cdot AE \cdot BF \cdot \frac{AC \cdot DE}{\text{AD}} =\]
\[= AB \cdot AE \cdot BF \cdot CD.\]
\[CD^{3} = AB \cdot AE \cdot BF.\]
\[\textbf{б)}\ CD^{2} =\]
\[= AD \cdot BD + теорема\ Пифагора:\ \]
\[AB^{2} = (AD + BD)^{2} =\]
\[= AD^{2} + BD^{2} + 2AD \cdot BD =\]
\[= AD^{2} + BD^{2} + 2CD^{2} =\]
\[= AE^{2} + BF^{2} + CD^{2} + 2CD^{2} =\]
\[= AE^{2} + BF^{2} + 3CD^{2}.\]
\[\textbf{в)}\ \frac{\text{AD}}{\text{AE}} = \frac{\text{AB}}{\text{AC}} \Longrightarrow AD = \frac{AB \cdot AE}{\text{AC}}.\]
\[\frac{\text{AC}}{\text{AD}} = \frac{\text{AD}}{\text{AE}} \Longrightarrow AD^{2} = AC \cdot AE.\]
\[AD^{3} = \frac{AB \cdot AE}{\text{AC}} \cdot AC \cdot AE =\]
\[= AB \cdot AE^{2}.\]
\[\frac{\text{AB}}{\text{BD}} = \frac{\text{BC}}{\text{BF}} \Longrightarrow \ BD = \frac{AB \cdot BF}{\text{BC}}.\]
\[\frac{\text{BC}}{\text{BD}} = \frac{\text{BD}}{\text{BF}} \Longrightarrow BD^{2} = BC \cdot BF.\]
\[BD^{3} = \frac{AB \cdot BF}{\text{BC}} \cdot BC \cdot BF =\]
\[= AB \cdot BF^{2}.\]
\[AB = AD + BD =\]
\[= \sqrt[3]{AB \cdot AE^{2}} + \sqrt[3]{AB \cdot BF^{2}}\ \ \ |\ :\sqrt[3]{\text{AB}}\]
\[\sqrt[3]{AB^{2}} = \sqrt[3]{AE^{2}} + \sqrt[3]{BF^{2}}.\]
\[Что\ и\ требовалось\ доказать.\]
