\[Дано:\]
\[ABCD - параллелограмм;\]
\[MNFK - прямоугольник;\]
\[AB = MN;\]
\[AD = NF;\]
\[S_{\text{ABCD}} = \frac{1}{2}S_{\text{MNFK}}.\]
\[Найти:\]
\[\angle BAD.\]
\[1)\ S_{\text{MNFK}} = MN \bullet FK = AB \bullet AD.\]
\[2)\ S_{\text{ABCD}} = AB \bullet AD \bullet \sin{\angle BAD}\]
\[AB \bullet AD \bullet \sin{\angle BAD} = \frac{1}{2}AB \bullet AD\]
\[\sin{\angle BAD} = \frac{1}{2}\text{\ \ \ }\]
\[\angle BAD = 30{^\circ}.\]
\[Ответ:\ \ 30{^\circ}.\]