\[\boxed{\mathbf{405.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - ромб;\]
\[AB = BD.\]
\[\mathbf{Найти:}\]
\[\textbf{а)}\ \angle A;\ \angle B;\ \angle C;\ \angle\text{D.}\]
\[\textbf{б)}\ \angle\text{ABD}\ и\ \angle\text{BAC.}\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ ABCD - ромб:\]
\[AB = BC = CD =\]
\[= AD\ (по\ определению).\]
\[2)\ \mathrm{\Delta}ABD\ и\ \mathrm{\Delta}BDC -\]
\[равносторонние:\]
\[\angle A = \angle ABD = \angle ADB = 60{^\circ};\]
\[\angle C = \angle CBD = \angle CDB = 60{^\circ}.\]
\[3)\ \angle B = \angle ABD + \angle DBC = 120{^\circ};\]
\[\angle D = \angle ADB + \angle BDC = 120{^\circ}.\]
\[\textbf{б)}\ Из\ свойства\ ромба:\]
\[\angle BAC = \frac{1}{2}\angle A = 30{^\circ};\ \]
\[\angle ABD = \frac{1}{2}\angle B = 60{^\circ}.\]
\[Ответ:а)\ \angle A = \angle C = 60{^\circ};\ \]
\[\angle B = \angle D = 120{^\circ};\ \]
\(б)\ \angle ABD = 60{^\circ};\ \angle BAC = 30{^\circ}.\)