Let $A_1, B_1, C_1$ be the feet of the altitudes through $A,B,C$. After applying the negative inversion centered at $H$, with radius $\sqrt{HA\cdot HA_1}=\sqrt{HB\cdot HB_1}=\sqrt{HC\cdot HC_1}$. The problem becomes: Let $I$ be the incenter of a triangle $\triangle DEF$. Let $T_1$ be the intersection of the perpendicular bisector of $ID$ and $EF$. Define $T_2,T_3$ similarly. Prove that: $T_1, T_2, T_3$ are collinear. Let $D_1$ be the midpoint of arc $EF$ not containing $D$. Define $D_2, D_3$ similarly. By incenter-excenter lemma $T_1=(E_1F_1) \cap (EF)$. Then by Desargues theorem, and since $(DD_1), (EE_1), (FF_1)$ are concurrent at $I$. $T_1,T_2$ and $T_3$ are collinear, as desired.

Let $O$ be the circumcenter of $ABC$ and $O_A$ be the circumcenter of $BHC$. It is well known that $AHO_AO$ is a parallelogram and hence $AT_a=AH=OO_A (1)$. Now using that $AO_A$ is the perpendicular bisector of $HT_a$ and $AHO_AO$ is a parallelogram we get that: $\angle O_AAT_a= \angle O_AAH= \angle AO_AO (2)$ From $ (1),(2)$ we get that $AT_aOO_A$ is isosceles trapezoid, hence $OT_a$ is parallel to $AO_A$, and hence perpendicular to $HT_a$. So $T_a$ lies on circle with diameter $OH$. Analogously $T_b$, $T_c$ lie on the same circle and so $H$ lies on the circumcircle of $T_aT_bT_c$, as needed.

Attachments: