ГДЗ по геометрии 8 класс Атанасян Задание 868

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

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[ABCD - параллелограмм;\]

\[N \in CD;\]

\[M = AN \cap BD;\]

\[P = AN \cap BC.\]

\[\mathbf{Доказать:}\]

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

\[\mathbf{Доказательство.}\]

\[1)\ Пусть\ AB = CD = a;\ \ \]

\[AD = BC = b.\]

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

\[(по\ двум\ углам):\]

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

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

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

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

\[(по\ двум\ углам):\]

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

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

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

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

\[(по\ теореме\ Фалеса):\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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