$A$ is the centroid of triangle $DBE$. Letting $F$ be the midpoint of $BD$, $2AF$ = $AE$ = $BD$ and so $F$ is the centre of circle $DAB$. So $\angle DAB$ is a right angle which implies the desired result.

Let $EA \cap DB=X$.By using Menelaus we get $DX=XB \implies A $ is centroid. Since $\frac{EA}{AX}=\frac{2}{1}$ and $AE=BD \implies DX=XA=XB$. So $\angle{DAB}=90$ and this implies the desired result