Let $ d$ be the distance from $ D$ to $ \overline{AB}$, and let $ e$ be the distance from $ E$ to $ \overline{AB}$. We have $ [BDE] \le [DEA]$, or $ \frac{1}{2} \cdot DE \cdot d \le \frac{1}{2} \cdot DE \cdot e \iff d\le e$. We also have $ [EAB]\le [ABD]$, or $ \frac{1}{2}\cdot AB \cdot e \le \frac{1}{2} \cdot AB \cdot d \iff e \le d$. Thus, $ d=e$, so $ ABDE$ is a trapezoid. Since it is also cyclic ($ \angle AEB = \angle ADB = 90^\circ$), it must be isosceles, so $ \angle EAB = \angle DBA$, and we are done.