\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[ABCD - прямоугольник;\]
\[\angle ABD = 64{^\circ}.\]
\[Найти:\]
\[\angle COD;\ \angle AOD.\]
\[Решение.\]
\[1)\ ABCD - прямоугольник:\]
\[AC = BD;\]
\[AO = OC;\]
\[BO = OD;\]
\[AO = \frac{1}{2}AC = \frac{1}{2}BD = BO;\]
\[AO = BO = CO = DO.\]
\[2)\ \mathrm{\Delta}AOB - равнобедренный:\]
\[\angle OAB = \angle OBA = 64{^\circ};\]
\[\angle OAB + \angle OBA + \angle AOB = 180{^\circ}\]
\[64{^\circ} + 64{^\circ} + \angle AOB = 180{^\circ}\]
\[\angle AOB = 52{^\circ}.\]
\[3)\ Сумма\ смежных\ углов:\]
\[\angle AOD + \angle AOB = 180{^\circ}\]
\[\angle AOD + 52{^\circ} = 180{^\circ}\]
\[\angle AOD = 128{^\circ}.\]
\[4)\ Вертикальные\ углы:\]
\[\angle COD = \angle AOB = 52{^\circ}.\]
\[Ответ:\ \ \angle COD = 52{^\circ};\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \angle AOD = 128{^\circ}.\]