let $W,X,Y,Z$ be centers of circles with diameters $AB,BC,CD,AD$ One sees that $WXYZ$ is a parallelogram lines $AA', BB', CC',DD'$ are radical axis of the pairs of circles, $WZ, WX, XY, YZ$ It follows that lines $AA', BB', CC', DD'$ are perpendicular to lines $WZ,WX,XY,YZ$ Then it follows that lines $A'D, B'C, C'B,D'A$ are parallel to lines $WZ, WX, XY, YZ$ But since lines $WZ,WX$ are parallel to lines $XY, YZ$ we get that $BC'$ is parallel to $A'D$ It also follows that $BA', DC'$ are parallel to lines $WZ,XY$ Therefore points $BA'C'D$ are collinear. it also follows that $CD'B'A$ angle A'B'C' = angle BB'C' - angle A'B'B =180- angle C'CB - angle A'B'B =90+ angle A'BA -angle C'CB = angle C'BC + angle A'BA = angle ABC