Решебник по геометрии 7 класс Атанасян ФГОС Задание 868

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Год:2020-2021-2022
Тип:учебник

Задание 868

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Геометрия 7 класс Атанасян ФГОС Просвещение
 
фгос Геометрия 7 класс Атанасян ФГОС, Бутузов Просвещение
Издание 1
Геометрия 7 класс Атанасян ФГОС Просвещение

\[\boxed{\mathbf{868.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - параллелограмм;\]

\[N \in CD;\]

\[M = AN \cap BD;\]

\[P = AN \cap BC.\]

\[\mathbf{Доказать:}\]

\[AM = \sqrt{MN \bullet MP}.\]

\[\mathbf{Доказательство.}\]

\[1)\ Пусть\ AB = CD = a;\ \ \]

\[AD = BC = b.\]

\[2)\ \mathrm{\Delta}BAM\sim\mathrm{\Delta}DNM\ \]

\[(по\ двум\ углам):\]

\[\frac{\text{AM}}{\text{MN}} = \frac{\text{AB}}{\text{ND}}\ \]

\[\frac{\text{AM}}{\text{MN}} = \frac{a}{a + NC}\]

\[\frac{\text{NC}}{a} + 1 = \frac{\text{MN}}{\text{AM}}.\]

\[3)\ \mathrm{\Delta}BMP\sim\mathrm{\Delta}DMA\ \]

\[(по\ двум\ углам):\]

\[\frac{\text{BP}}{\text{AD}} = \frac{\text{MP}}{\text{AM}};\ \ \]

\[MP = \frac{\text{BP}}{b}AM = \frac{b - PC}{b}AM =\]

\[= \left( 1 - \frac{\text{PC}}{b} \right)\text{AM.}\]

\[4)\ PC \parallel AD\ \]

\[(по\ теореме\ Фалеса):\]

\[\frac{\text{AP}}{\text{NP}} = \frac{\text{CD}}{\text{NC}} = \frac{a}{\text{NC}};\ \ \ \]

\[\frac{\text{AP}}{\text{NP}} = \frac{a}{\text{NC}};\ \ \]

\[\ \frac{\text{NC}}{a} = \frac{\text{NP}}{\text{AP}};\]

\[\frac{\text{PC}}{\text{AD}} = \frac{\text{NP}}{\text{AN}};\ \ \ \]

\[\frac{\text{PC}}{b} = \frac{\text{NP}}{\text{AN}}.\]

\[5)\ MP = \left( 1 - \frac{\text{PC}}{b} \right)AM =\]

\[= \left( 1 - \frac{\text{NP}}{\text{AN}} \right)AM =\]

\[= \frac{AN - NP}{\text{AN}}AM = \frac{\text{AP}}{\text{AN}}\text{AM.}\]

\[\frac{\text{NP}}{\text{AP}} + 1 = \frac{\text{MN}}{\text{AM}};\ \ \ \]

\[\frac{NP + AP}{\text{AP}} = \frac{\text{MN}}{\text{AM}};\ \ \ \]

\[\frac{\text{AN}}{\text{AP}} = \frac{\text{MN}}{\text{AM}}\]

\[\frac{\text{MN}}{\text{AM}} = \frac{\text{AN}}{\text{AP}} = \frac{\text{AM}}{\text{MP}}\]

\[AM^{2} = MN \bullet MP\]

\[AM = \sqrt{MN \bullet MP}.\]

\[\mathbf{Что\ и\ требовалось\ доказать.}\]

Издание 2
фгос Геометрия 7 класс Атанасян ФГОС, Бутузов Просвещение

\[\boxed{\mathbf{868.еуроки - ответы\ на\ пятёрку}}\]

\[\mathbf{Дано:}\]

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

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

\[CD\bot AB;\]

\[DE\bot AC;\]

\[DF\bot BC.\]

\[\mathbf{Доказать:}\]

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

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

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

\[\mathbf{Доказательство.}\]

\[Подобные\ треугольники,\ \]

\[полученные\ в\ результате\ \]

\[построения:\ \]

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

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

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

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

\[CD =\]

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

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

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

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

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

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

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

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

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

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

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

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

\[\textbf{б)}\ CD^{2} =\]

\[= AD \cdot BD + теорема\ Пифагора:\ \]

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

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

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

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

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

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

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

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

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

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

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

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

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

\[AB = AD + BD =\]

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

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

\[Что\ и\ требовалось\ доказать.\]

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