\[\boxed{\mathbf{881.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[Окружность\ (O;R);\]
\[OA = R;\]
\[BD - касатель;\]
\[OB\bot BD;\]
\[AD\bot BD.\]
\[\mathbf{Доказать:}\]
\[\frac{AB^{2}}{\text{AD}} = const.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ \text{A\ }не\ лежит\ на\ \]
\[диаметре,\ проведем\ диаметр\ \]
\[\text{BC.}\ \]
\[2)\ \angle ABD = \frac{1}{2} \cup AB = \angle ACB.\]
\[3)\ \mathrm{\Delta}BAC - прямоугольный,\ \]
\[так\ как\ \angle\text{CAB\ }опирается\ на\ \]
\[диаметр:\]
\[\angle A = 90{^\circ};\]
\[4)\ \mathrm{\Delta}\text{ADB\ }и\ \mathrm{\Delta}BAC -\]
\[прямоугольные;\ \ \mathrm{\Delta}ADB\sim\mathrm{\Delta}BAC\ \]
\[(по\ двум\ углам):\]
\[\angle ABD = \angle BCA.\ \]
\[Отсюда:\]
\[\frac{\text{AD}}{\text{AB}} = \frac{\text{AB}}{\text{BC}}\]
\[\frac{AB^{2}}{\text{AD}} = BC = 2r.\]
\[5)\ A\ лежит\ на\ диаметре \Longrightarrow\]
\[\Longrightarrow AD = AB = 2r:\]
\[\frac{AB^{2}}{\text{AD}} = 2r.\]
\[6)\ \frac{AB^{2}}{\text{AD}} = 2r \Longrightarrow \ постоянно\ \]
\[равно\ диаметру.\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{881.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[N \in CD;\]
\[M = AN \cap BD;\]
\[P = AN \cap BC.\]
\[\mathbf{Доказать:}\]
\[AM = \sqrt{MN \bullet MP}.\]
\[\mathbf{Доказательство.}\]
\[1)\ Пусть\ AB = CD = a;\ \]
\[\ AD = BC = b.\]
\[2)\ \mathrm{\Delta}BAM\sim\mathrm{\Delta}DNM\ (по\ двум\ углам):\]
\[\frac{\text{AM}}{\text{MN}} = \frac{\text{AB}}{\text{ND}}\ \]
\[\frac{\text{AM}}{\text{MN}} = \frac{a}{a + NC}\]
\[\frac{\text{NC}}{a} + 1 = \frac{\text{MN}}{\text{AM}}.\]
\[3)\ \mathrm{\Delta}BMP\sim\mathrm{\Delta}DMA\ (по\ двум\ углам):\]
\[\frac{\text{BP}}{\text{AD}} = \frac{\text{MP}}{\text{AM}};\ \ \]
\[MP = \frac{\text{BP}}{b}AM = \frac{b - PC}{b}AM =\]
\[= \left( 1 - \frac{\text{PC}}{b} \right)\text{AM.}\]
\[4)\ PC \parallel AD\ (по\ теореме\ Фалеса):\]
\[\frac{\text{AP}}{\text{NP}} = \frac{\text{CD}}{\text{NC}} = \frac{a}{\text{NC}};\ \ \ \]
\[\frac{\text{AP}}{\text{NP}} = \frac{a}{\text{NC}};\ \ \]
\[\ \frac{\text{NC}}{a} = \frac{\text{NP}}{\text{AP}};\]
\[\frac{\text{PC}}{\text{AD}} = \frac{\text{NP}}{\text{AN}};\ \ \ \]
\[\frac{\text{PC}}{b} = \frac{\text{NP}}{\text{AN}}.\]
\[5)\ MP = \left( 1 - \frac{\text{PC}}{b} \right)AM =\]
\[= \left( 1 - \frac{\text{NP}}{\text{AN}} \right)AM =\]
\[= \frac{AN - NP}{\text{AN}}AM = \frac{\text{AP}}{\text{AN}}\text{AM.}\]
\[\frac{\text{NP}}{\text{AP}} + 1 = \frac{\text{MN}}{\text{AM}};\ \ \ \]
\[\frac{NP + AP}{\text{AP}} = \frac{\text{MN}}{\text{AM}};\ \ \ \]
\[\frac{\text{AN}}{\text{AP}} = \frac{\text{MN}}{\text{AM}}\]
\[\frac{\text{MN}}{\text{AM}} = \frac{\text{AN}}{\text{AP}} = \frac{\text{AM}}{\text{MP}}\]
\[AM^{2} = MN \bullet MP\]
\[AM = \sqrt{MN \bullet MP}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]