imc

First, let's call the center of both squares $I$. Then, $\angle AIE=20$, and since $\overline{EI}=\overline{AI}$, $\angle AEI=\angle EAI=80$. Then, we know that $AI$ bisects angle $\angle DAB$, so $\angle BAI=\angle DAI=45$. Subtracting $45$ from $80$, we get $\boxed{35^{\circ }}$