ILL 1974 Problem 46

Outside an arbitrary triangle $ABC$ , triangles $ADB$ and $BCE$ are constructed such that $\angle ADB=\angle BEC=90^{\circ}$ and $\angle DAB=\angle EBC=30^{\circ}$ . On the segment $AC$ the point $F$ with $AF=3FC$ is chosen. Prove that $\angle DFE=90^{\circ}$ and $\angle FDE=30^{\circ}$ .