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$H_i$ lies on the ray $IA_i$ and $IH_i\cdot IA_i=2r^2$ (where $r=IB_i$), since the midpoint of $IH_i$ is the midpoint of $B_iB_{i-1}$. Hence $H_i$ is the image of $A_i$ under the inversion with respect to center $I$ and radius $\sqrt{2}r$. Thus the result.