$\angle BXD = 135-(\angle B+\angle C)/2-\angle YCA=\angle CYD$

Nice angle chasing problem! $\angle ABX=\angle AXY=\angle AYX=\angle ACY$, then we can easily have $AE*AC=AY^2=AX^2=AF*AB$ then $(B F E C)$ is cyclic and $\angle XYD=\angle XAD/2=90-\angle ADX=\angle XDB$,which gives that $BC$ is tangent to $(X D Y)$. Let intersection of $DY$ and $AC$ is $T$, and intersection $DX$ and $AB$ is $P$. Then $\angle YTC= \angle YAE+\angle AYE+\angle EYT= \angle AEF+\angle XYD=\angle FBC+\angle XDB$= $\angle XPB$ $(1)$. And because of $AY=AX$ we have $\angle XBP=\angle YCT$ $(2)$. $(1)$ and $(2)$ together completes the problem.