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МатематикаГДЗ по геометрии 9 класс Атанасян ФГОС Задание 912
Задание 912
\[\boxed{\mathbf{912.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[BD \cap AC = O;\]
\[AM = MO;\]
\[M \in AO.\]
\[\mathbf{Найти:}\]
\[k - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ \overrightarrow{\text{AC}} = k\overrightarrow{\text{AO}};\ \overrightarrow{\text{AC}} \uparrow \uparrow \overrightarrow{\text{AO}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{AC}} \right| = 2\left| \overrightarrow{\text{AO}} \right| \Longrightarrow k = 2;\]
\[\textbf{б)}\ \overrightarrow{\text{BO}} = k\overrightarrow{\text{BD}};\ \overrightarrow{\text{BO}} \uparrow \uparrow \overrightarrow{\text{BD}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{BO}} \right| = \frac{1}{2}\left| \overrightarrow{\text{BD}} \right| \Longrightarrow\]
\[\Longrightarrow k = \frac{1}{2};\]
\[\textbf{в)}\ \overrightarrow{\text{OC}} = k\overrightarrow{\text{CA}};\ \overrightarrow{\text{OC}} \uparrow \downarrow \overrightarrow{\text{CA}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{OC}} \right| = \frac{1}{2}\left| \overrightarrow{\text{CA}} \right| \Longrightarrow\]
\[\Longrightarrow k = - \frac{1}{2};\]
\[\textbf{г)}\ \overrightarrow{\text{AB}} = k\overrightarrow{\text{DC}};\ \overrightarrow{\text{AB}} \uparrow \uparrow \overrightarrow{\text{DC}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{AB}} \right| = 1\left| \overrightarrow{\text{DC}} \right| \Longrightarrow\]
\[\Longrightarrow k = 1;\]
\[\textbf{д)}\ \overrightarrow{\text{BC}} = k\overrightarrow{\text{DA}};\ \overrightarrow{\text{BC}} \uparrow \downarrow \overrightarrow{\text{DA}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{BC}} \right| = 1\left| \overrightarrow{\text{DA}} \right| \Longrightarrow\]
\[\Longrightarrow k = - 1;\]
\[\textbf{е)}\ \overrightarrow{\text{AM}} = k\overrightarrow{\text{CA}};\ \overrightarrow{\text{AM}} \uparrow \downarrow \overrightarrow{\text{CA}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{AM}} \right| = \frac{1}{4}\left| \overrightarrow{\text{CA}} \right| \Longrightarrow\]
\[\Longrightarrow k = - \frac{1}{4};\]
\[\textbf{ж)}\ \overrightarrow{\text{MC}} = k\overrightarrow{\text{AM}};\ \overrightarrow{\text{MC}} \uparrow \uparrow \overrightarrow{\text{AM}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{MC}} \right| = 3\left| \overrightarrow{\text{AM}} \right| \Longrightarrow\]
\[\Longrightarrow k = 3;\]
\[\textbf{з)}\ \overrightarrow{\text{AC}} = k\overrightarrow{\text{CM}};\ \overrightarrow{\text{AC}} \uparrow \downarrow \overrightarrow{\text{CM}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{AC}} \right| = \frac{4}{3}\left| \overrightarrow{\text{CM}} \right| \Longrightarrow\]
\[\Longrightarrow k = - \frac{4}{3};\]
\[\textbf{и)}\ \overrightarrow{\text{AO}} = k:\ \]
\[\overrightarrow{\text{AO}}\ и\ \overrightarrow{\text{BD}} - неколлинеарны \Longrightarrow\]
\[\Longrightarrow их\ нельзя\ разложить.\]
\[\boxed{\mathbf{912.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC - равносторонний;\]
\[X \in \cup AB.\]
\[\mathbf{Доказать:}\]
\[XC = XA + XB.\]
\[\mathbf{Доказательство.}\]
\[1)\ Отложим\ на\ отрезке\ \text{XC\ }\]
\[отрезок\ XD = XA.\]
\[2)\ \mathrm{\Delta}ADX - равнобедренный:\]
\[\angle AXD = \angle ABC = \frac{1}{2} \cup AC = 60{^\circ};\]
\[\mathrm{\Delta}AXD - равносторонний;\]
\[AD = XA;\]
\[\ \angle ADX = \angle DAX = 60{^\circ}.\]
\[3)\ \mathrm{\Delta}ADC = \mathrm{\Delta}AXB - по\ двум\ \]
\[сторонам\ и\ углу\ между\ ними:\]
\[AC = AB;\ \]
\[AD = AX;\]
\[\angle CAD = \angle BAX = 60{^\circ} - \angle DAB.\]
\[Отсюда:\ \ DC = XB.\]
\[4)\ XC = XD + DC = XA + XB.\]
\[Что\ и\ требовалось\ доказать.\]
