[asy][asy] size(8cm); defaultpen(fontsize(8pt)); pen pri=green; pen sec=springgreen; pen tri=chartreuse; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A, B, C, H, K, Ic, X, Y; A=(0,0); B=(1,1); C=(3,0); H=foot(B, A, C); K=(3-sqrt(5),0); Ic=circumcenter(A,B,K); X=intersectionpoint(Ic--C,B--K); Y=intersectionpoint(Ic--H,A--B); draw(A--B--C--cycle,pri); draw(B--H,sec); draw(B--K,sec); draw(Ic--C,tri+dashed); draw(Ic--H,tri+dashed); dot("$A$",A,W); dot("$B$",B,N); dot("$C$",C,E); dot("$K$",K,S); dot("$H$",H,S); dot("$I_C$",Ic,N); dot("$X$",X,NW); dot("$Y$",Y,S); [/asy][/asy] Since $\angle A = 45^{\circ}$ and $\angle AHB = 90^{\circ}$, $\angle B = 45^{\circ}$. Thus $\Delta ABH$ is an isosceles right triangle. Draw the internal angle bisector of $\angle C$ and the external angle bisector of $\angle BHC$. These will intersect at the $C$-excenter of $\Delta BHC$. Our goal is to show that both of these lines are also perpendicular bisectors of the sides of $\Delta ABK$. Let $X = I_CC \cap BK$ and $Y = I_CH \cap AB$. It is clear that $Y$ is the midpoint of $\overline{AB}$, since $\overline{I_CH}$ is the external angle bisector of $\angle BHC$ and thus the internal angle bisector of $\angle AHB$. Since the internal angle bisector of an isosceles triangle is its altitude and its median, $\overline{HY}$ is a perpendicular bisector of $\overline{AB}$. A similar result follows for point $X$ (it is the foot of the altitude from $C$ to $\overline{BK}$, and we are given that $BC = CK$.) Thus, both the internal angle bisector of $\angle C$ and the external angle bisector of $\angle BHC$ are perpendicular bisectors of $\Delta ABK$, and it follows that the center of the circumscribed circle of $\Delta ABK$ coincides with the center of an excircle of $\Delta BCH$. $\square$