ГДЗ по геометрии 9 класс Атанасян Задание 537

\[\boxed{\mathbf{537.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

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

\[\angle\text{BAD} = \angle\text{DAC};\]

\[\text{AB} = 14\ см;\]

\[\text{BC} = 20\ см;\]

\[\text{AC} = 21\ см.\]

\[\mathbf{Найти:}\]

\[\text{BD} - ?\]

\[\text{DC} - ?\]

\[\mathbf{Решение.}\]

\[1)\ \text{AH} - высота,\ общая\ для\ \]

\[\mathrm{\Delta}\text{ABD}\ и\ \mathrm{\Delta}\text{ACD}:\]

\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{BD}}{\text{CD}}.\]

\[2)\ \angle\text{BAD} = \angle\text{DAC}:\]

\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{AB} \bullet \text{AD}}{\text{AD} \bullet \text{AC}} = \frac{\text{AB}}{\text{AC}}.\]

\[3)\frac{\text{AB}}{\text{AC}} = \frac{\text{BD}}{\text{CD}}\]

\[\text{AC} \bullet \text{BD} = \text{AB} \bullet \text{CD}\]

\[\frac{\text{BD}}{\text{DC}} = \frac{\text{AB}}{\text{AC}}.\]

\[4)\ Пусть\ \text{BD} = x;\]

\[\text{DC} = \text{BC} - x = 20 - x.\]

\[5)\frac{x}{20 - x} = \frac{14}{21}\]

\[\frac{x}{20 - x} = \frac{2}{3}\]

\[3x = 2(20 - x) = 40 - 2x\]

\[5x = 40\]

\[x = 8\ см.\]

\[3)\ \text{DC} = 20 - 8 = 12\ см.\]

\[\mathbf{Ответ:}\text{BD} = 8\mathbf{\ см};\text{DC} = 12\ см.\]