\[Схематический\ рисунок.\]
\[Дано:\]
\[ABCD - равнобокая\ трапеция;\]
\[\angle DCE = 65{^\circ};\]
\[\angle D = 35{^\circ};\]
\[CE \parallel AB.\]
\[Найти:\]
\[\angle A;\ \angle B;\ \]
\[\angle C;\ \angle D.\]
\[Решение.\]
\[1)\ ABCD - трапеция:\]
\[BC \parallel AD;\ \ \ \]
\[AB \nparallel CD.\]
\[2)\ \mathrm{\Delta}ECD:\]
\[\angle DCE + \angle DEC + \angle EDC = 180{^\circ}\]
\[65{^\circ} + \angle DEC + 35{^\circ} = 180{^\circ}\]
\[\angle DEC = 80{^\circ}.\]
\[3)\ BCEA - четырехугольник:\]
\[BC \parallel AE;\ \ \ \]
\[AB \parallel CE.\]
\[BCEA - параллелограмм:\]
\[\angle AEC = 180{^\circ} - \angle DEC = 100{^\circ};\]
\[\angle B = \angle E = 100{^\circ};\ \ \]
\[\angle A = \angle C.\]
\[\angle A + \angle B + \angle C + \angle E = 360{^\circ}\]
\[\angle A + 100{^\circ} + \angle A + 100{^\circ} = 360{^\circ}\]
\[2\angle A = 160{^\circ}\ \ \ \]
\[\angle C = \angle A = 80{^\circ}.\]
\[4)\ \angle BCD = \angle BCE + \angle DCE\]
\[\angle BCD = 80{^\circ} + 65{^\circ} = 145{^\circ}.\]
\[Ответ:\ \ 80{^\circ};\ 100{^\circ};\ 145{^\circ};\ 35{^\circ}.\]