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МатематикаРешебник по геометрии 8 класс Атанасян ФГОС Задание 843
Задание 843
\[\boxed{\mathbf{843.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\ \ \]
\[D \in BA;\ \ AD = DC;\]
\[K \in BA;M \in BC;\]
\[S_{\text{BDM}} = S_{\text{BCK}};\]
\[\angle BAC = \alpha.\]
\[\mathbf{Найти:}\]
\[\angle BKM - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{BDM}} = S_{\text{BCK}};\ \ \angle B - общий:\]
\[BD \bullet BM = BK \bullet BC.\ \]
\[Отсюда:\]
\[\frac{\text{BD}}{\text{BK}} = \frac{\text{BC}}{\text{BM}}\ (\angle B - общий);\ \ \]
\[\mathrm{\Delta}BKM\sim\mathrm{\Delta}BDC.\]
\[Значит:\]
\[DC \parallel KM;\ \ \]
\[\angle BKM = \angle BDC = \angle ADC.\]
\[2)\ AD = AC \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ADC - равнобедренный;\]
\[DC - основание.\ \]
\[Следовательно:\]
\[\angle ADC = \frac{1}{2} \bullet (180 - \angle DAC) = \frac{1}{2}\text{α.}\]
\[\angle BKM = \angle ADC = \frac{1}{2}\text{α.}\]
\[Ответ:\frac{1}{2}\text{α.}\]
\[\boxed{\mathbf{843.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[M \in AC;K \in BC;\]
\[O = AK \cap BM;\]
\[S_{\text{OMA}} = S_{1};\ \]
\[S_{\text{OAB}} = S_{2};\]
\[S_{\text{OBK}} = S_{3}.\]
\[\mathbf{Найти:}\]
\[S_{\text{CMK}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ Проведем\ несколько\ \]
\[перепендикуляров:\]
\[\text{OD}\bot AC;\ \ BE\bot AC;KF\bot AC;\]
\[OG\bot BC;\ \ AP\bot BC;MH\bot BC.\]
\[2)\ Площади\ треугольников:\]
\[\ S_{\text{MOK}} = S_{4};\ \ S_{\text{ABC}} = S.\ \]
\[3)\ S_{\text{CMK}} =\]
\[= S - \left( S_{1} + S_{2} + S_{3} + S_{4} \right).\]
\[Преобразуем:\]
\[\frac{S_{\text{AMK}}}{S_{\text{AMB}}} = \frac{S_{1} + S_{4}}{S_{1} + S_{2}} = \frac{\text{KF}}{\text{BE}} =\]
\[= \frac{\text{KF}}{\text{BE}} \bullet \frac{\text{MC}}{\text{MC}} = \frac{S_{\text{CMK}}}{S - \left( S_{1} + S_{2} \right)} =\]
\[= \frac{S_{\text{CMK}}}{S_{3} + S_{4} + S_{\text{CMK}}}.\]
\[\frac{S_{3}}{S_{2}} = \frac{\text{OK}}{\text{OA}} = \frac{S_{4}}{S_{1}};\ \ \ \]
\[S_{4} = \frac{S_{1}S_{3}}{S_{2}}.\]
\[Решим\ уравнение\ \]
\[относительно\ S_{\text{CMK}}:\]
\[\left( S_{1} + \frac{S_{1}S_{3}}{S_{2}} \right)\left( S_{3} + \frac{S_{1}S_{3}}{S_{2}} + S_{\text{CMK}} \right) =\]
\[= \left( S_{1} + S_{2} \right) \bullet S_{\text{CMK}}\]
\[\left( S_{1} + \frac{S_{1}S_{3}}{S_{2}} \right)\left( S_{3} + \frac{S_{1}S_{3}}{S_{2}} \right) =\]
\[= (S_{1} + S_{2} - S_{1}) \bullet \left( 1 + \frac{S_{3}}{S_{2}} \right)S_{\text{CMK}}\]
\[S_{\text{CMK}} =\]
\[= \frac{\left( S_{1} + \frac{S_{1}S_{3}}{S_{2}} \right)\left( S_{3} + \frac{S_{1}S_{3}}{S_{2}} \right)}{S_{1} + S_{2} - S_{1}\left( 1 + \frac{S_{3}}{S_{2}} \right)} =\]
\[= \frac{S_{1}S_{3}\frac{\left( S_{2} + S_{3} \right)}{S_{2}^{2}}}{\frac{S_{2}\left( S_{1} + S_{2} \right) - S_{1}\left( S_{2} + S_{1} \right)}{S_{2}}} =\]
\[= \frac{S_{1}S_{3}\left( S_{1} + S_{2} \right)\left( S_{2} + S_{3} \right)}{S_{2}\left( S_{1}S_{2} + S_{2}^{2} - S_{1}S_{2} - S_{1}S_{3} \right)} =\]
\[= \frac{S_{1}S_{3}\left( S_{1} + S_{2} \right)\left( S_{2} + S_{3} \right)}{S_{2}(S_{2}^{2} - S_{1}S_{3})}.\]
\[Ответ:S_{\text{CMK}} =\]
\[= \frac{S_{1}S_{3}\left( S_{1} + S_{2} \right)\left( S_{2} + S_{3} \right)}{S_{2}(S_{2}^{2} - S_{1}S_{3})}.\]
