$A$ is the $C-$excenter of $\triangle CFE$. Remark: there is no need of the value of $\widehat{BAE}$ Best regards, sunken rock

Let the altitude from $A$ to $EF$ be $G$, and extend $BC$ from $B$ to a point $K$ so that $BK = DF$ and connect $AK$. Notice that since $\angle KBA = \angle FDA = 90$ and $AB = AD$ we must have $\triangle KBA \cong \triangle FDA$ so $KA = FA$. We also have $\angle KAE = \angle FAE = 45$ and $AE = AE$ so $\triangle KAE \cong \triangle FAE$. Since $AB$ and $AG$ are corresponding altitudes, we must have $AB = AG = 1$ and we are done.