\[\boxed{\mathbf{438.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\text{ABCD} - трапеция;\]
\[\text{AC}\bot\text{CD};\]
\[\angle\text{BAC} = \angle\text{CAD};\]
\[\angle D = 60{^\circ};\]
\[P_{\text{ABCD}} = 20\ см.\]
\[\mathbf{Найти:}\]
\[\text{AD} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \angle\text{CAD} = 90{^\circ} - 60{^\circ} = 30{^\circ}\]
\[По\ свойству\ прямоугольного\ \]
\[треугольника:\]
\[\text{CD} = \frac{1}{2}\text{AD}\text{.\ }\]
\[2)\ \angle\text{BAC} = \angle\text{CAD} = 30{^\circ}.\]
\[3)\ \angle A = \angle\text{BAC} + \angle\text{CAD} =\]
\[= 30 + 30 = 60{^\circ}\]
\[\angle A = \angle D = 60{^\circ}.\]
\[Следовательно:\]
\[\text{ABCD} - равнобедренная\ \]
\[трапеци \Longrightarrow \text{AB} = \text{CD}.\]
\[4)\ \text{ABCD} - трапеция \Longrightarrow\]
\[\Longrightarrow \text{BC} \parallel \text{AD}.\]
\[5)\ \text{BC} \parallel \text{AD}\ и\ \text{AC} - секущая:\ \]
\[\angle\text{BCA} = \angle\text{CAD} =\]
\[= 30{^\circ}\ (как\ накрестлежащие).\]
\[6)\ \angle\text{BAC} = \angle\text{BCA} = 30{^\circ}:\]
\[\mathrm{\Delta}\text{ABC} - равнобедренный \Longrightarrow\]
\[\Longrightarrow \ \text{AB} = \text{BC}.\]
\[7)\ P_{\text{ABCD}} =\]
\[= \text{AB} + \text{BC} + \text{CD} + \text{AD} = 20\ см;\]
\[\text{AB} = \text{BC} = \text{CD};\]
\[P_{\text{ABCD}} = 3\text{AB} + \text{AD} = 20\ см;\]
\[\text{CD} = \frac{1}{2}\text{AD};\]
\[P_{\text{ABCD}} = 3 \bullet \frac{1}{2}\text{AD} + \text{AD} = 20\ см;\]
\[\frac{5}{2}\text{AD} = 20\ см\]
\[\text{AD} = 8\ см.\]
\[Ответ:\text{AD} = 8\ см.\]