First note that, we must have $\angle BCA = \angle CAB = 45^\circ$. Then, $\angle BDA = 45^\circ$ which gives us that $\angle CDF = 180 - 45-60 =75^\circ$. Now, $\measuredangle BDA = 90^\circ$ ($DA=DB$ since $D$ is the center of $(ABC)$). This means, $\angle FDB = 180 - 45 - 45-75=15^\circ$. Now, note that,
\[EB=AE=ED=EF\]Thus, $E$ is the center of a circle passing through points $A,D,F$ and $B$. This means, $ADFB$ is cyclic and thus, $\angle FAB = 15^\circ$. This then, gives $\angle FAC = 30^\circ$ which in turn implies $\angle FBD = 30^\circ$. Now,
\[\frac{DX}{DB} = \tan(30^\circ)=\frac{DY}{AD}\]Further,
\[\frac{DX}{DY}=\frac{DB}{AD}=1\](since $D$ is the center of $(ABC)$). This means indeed $DX=DY$ as was required.