\[Схематический\ рисунок.\]
\[Дано:\]
\[ABCD - трапеция;\]
\[AB = BC = CD;\]
\[AC\bot CD.\]
\[Найти:\]
\[\angle A;\ \angle B;\ \angle C;\ \angle D.\]
\[Решение.\]
\[1)\ ABCD - трапеция:\]
\[BC \parallel AD;\ \ \]
\[AB = CD;\]
\[\angle A = \angle D;\ \ \]
\[\angle B = \angle C.\]
\[2)\ \mathrm{\Delta}ABC - \ равнобедренный:\]
\[\angle BCA = \angle BAC.\]
\[3)\ Для\ AD\ и\ \text{BC\ }и\ секущей\ AC:\]
\[\angle DAC = \angle BCA = \angle BAC.\]
\[4)\ \mathrm{\Delta}ACD - прямоугольный:\]
\[\angle BAD = 2\angle DAC;\]
\[\angle ADC = \angle BAD = 2\angle DAC;\]
\[\angle DAC + \angle ADC = 90{^\circ}\]
\[\angle DAC + 2\angle DAC = 90{^\circ}\]
\[3\angle DAC = 90{^\circ}\ \ \]
\[\angle DAC = 30{^\circ}.\]
\[\angle ADC = 2 \bullet 30{^\circ} = 60{^\circ}.\]
\[5)\ ABCD - трапеция:\]
\[\angle A + \angle B = 180{^\circ}\]
\[60{^\circ} + \angle B = 180{^\circ}\]
\[\angle B = 120{^\circ}.\]
\[Ответ:\ \ 60{^\circ};\ 120{^\circ};\ 120{^\circ};\ 60{^\circ}.\]