, 2023

Let $M$, $N$ be the midpoint of minor arc and major arc $BC$ of circle $(O)$, respectively, then points $A$, $D$, $M$ are collinear. Let $T$ be the intersection of the two tangent lines at $B$ and $C$ of $(O)$ and $J$ the midpoint of $BC$. First, by the angle chasing, we have $\angle TBM = \angle BAM = \angle MBC$ so $BM$ is the internal bisector of $\angle CBT$. Thus, $M$ is the center of the incircle of the triangle $BTC$, denoted by $(M)$. Then it is easy to see that $D$ is the internal homothety center of $(\omega)$ and $(M)$; and $T$ is the external homothety center of $(M)$ with $(O)$.

Applying Monge D'Alembert's theorem to three circles $(I)$, $(M)$, $(O)$, we see that $U$, $D$, $T$ are collinear which implies that $T$ belongs to the fixed line $TD$.

Let $V$ be the external homothety center of $(\omega)$ with $(O)$, since $IH \parallel MN$, points $V$, $H$, $M$ are collinear. Note that since $U$ is the inner center of $(\omega)$ and $(O)$, $H$, $U$, $N$ is collinear because $IH \parallel ON$. Then we have
$$A(UH, IN) = A(UM, HN) = M(UH, AN) = M(UV, IO).$$
On the other hand, since $U$, $V$ is the internal, external homothety centers of $(\omega)$ and $(O)$, $(UV, IO) = -1$. From this it follows that $A(UH, IN) = -1$. Since $AI \perp AN$, by the property of harmonic bundles, $AI$ is the internal bisector of angle $HAN$. $\square$