Zooming in, we can repat the reflection and get the blue triangle. Its circle of circumference is the famous Feulerbach (nine point) circle.
It is the image of the brown circle. Thus its center U** it the image of U*. With a bit of computation, we see that U** bisects HU*.
Furthermore, look at the fat rectangle. AF is mapped to A*F*. The brown height is also the middle perpendicular of the green triangle. Thus it contains U*. By the above remark U** lies on the middle line of the rectangle. Thus U**F=U**A* and F is the base point of the height in the green triangle. F lies on the Feurbach circle.
We now found six points on the Feuerbach circle, the middle points on the sides and the base points of the heights. Now the blue circle has half the radius of the brown one, since it is the image of it. So 2U**X=U*A. Furthermore we know that 2HU**=HU*. So HX=XA. So the other three points are the middle points of HA, HB, HC.