The condition $\angle ABC = 45^\circ$ is irelevant, it is sufficient (and necessary) that the angle $\angle CAB = 30^\circ.$ This makes the triangle $\triangle COB$ equilateral. Let q be a distance greater than or equal to the distance of the circumcenter O from the side CA. Points $X, X' \in CA$ such that OX = OX' = q are intersections of CA with a circle $\mathcal C_1$ centered at O and with radius q, while points $Y, Y' \in BC$ such that BY = BY' = q are intersections of BC with a circle $\mathcal C_2$ centered at B and with the same radius q. Since the circles $\mathcal C_1 \cong \mathcal C_2$ are congruent and since CB = CO, powers of the point C to these 2 circles are equal, $CX \cdot CX' = CY \cdot CY'.$ This means that the quadrilateral XX'YY' is cyclic. The perpendicular bisector of the side YY' is fixed, namely a normal to BC at B, and the perpendicular bisector of the side XX' is also fixed, identical with the perpendicular bisector of CA. Thus the circumcenter P of the cyclic quadrilateral XX'YY' is fixed and the perpendicular bisectors of its remaining sides XY', X'Y and of its diagonals XY, X'Y' all pass through its fixed circumcenter P.
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