\[Схематический\ рисунок.\]
\[Дано:\]
\[ABCD - трапеция;\]
\[AB = CD;\]
\[\angle BAC = 20{^\circ};\]
\[\angle CAD = 50{^\circ}.\]
\[Найти:\]
\[\angle ACB;\ \angle ACD.\]
\[Решение.\]
\[1)\ ABCD - равнобокая\ трапеция:\]
\[AB = CD;\ \ \ \]
\[AD \parallel BC;\]
\[\angle A = \angle C;\ \ \]
\[\angle B = \angle D.\]
\[2)\ Для\ AD\ и\ \text{BC\ }и\ секущей\ AC:\]
\[\angle ACB = \angle CAD = 50{^\circ}.\]
\[3)\ Для\ AD\ и\ \text{BC\ }и\ секущей\ AB:\]
\[\angle A = \angle CAD + \angle BAC = 70{^\circ}\]
\[\angle DAB + \angle ABC = 180{^\circ}\]
\[70{^\circ} + \angle ABC = 180{^\circ}\]
\[\angle ABC = 110{^\circ}.\]
\[4)\ \angle C = \angle B = 110{^\circ};\]
\[\angle DCB = \angle ACB + \angle ACD\]
\[110{^\circ} = 50{^\circ} + \angle ACD\]
\[\angle ACD = 60{^\circ}.\]
\[Ответ:\ \ \angle ACB = 50{^\circ};\ \angle ACD = 60{^\circ}.\]