From Pythagorean theorem on triangle $ABC$ we have $AC=\sqrt{2}AB$ and we have $\angle BCA=45^{\circ}$. So, we have $\angle ACD=90^{\circ}$. Again from Pythagorean theorem on triangle $ACD$ we have $AD=\sqrt{3}AB$. Now let $F$ be foot of triangle $DEA$ which is on side $EA$. Then triangle $DEF$ has angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ so it is know that $DF=\frac{\sqrt{3}}{2}DE=\frac{\sqrt{3}}{2}AB$. So, on triangle $DFA$ we have $\angle DFA=90^{\circ}$ and $AD=2DF$, from here it is known that the other two angles are $\angle ADF=60^{\circ}$ and $\angle DAF=30^{\circ}$. Finally we have $\angle DAE=\angle DAF=30^{\circ}$.

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