I described my solution in the case of an acute triangle, but it is also true in the case of an obtuse triangle with some changes: some external bisectors will become internal, some will internal become external. I will show that all these lines are parallel to $OH$. Note that $BC$, $AC$ are the external bisectors of the triangle $BC_2A$, so $CC_2$ - bisectors of $\angle AC_2B$. $\angle AC_2B = 180^{\circ} - \angle C_2AB - \angle C_2BA = 180^{\circ} - (180^{\circ} - 2\angle A) - (180^{\circ} - 2\angle B) = 180^{\circ} - 2\angle C$ so $C_2ABO$ is cyclic. But $BO = OA = R$ so $O$ lie on the bisector of $\angle AC_2B$ . Sum up, $C_2, O, C$ are collinear. Let I be the center of the inscribed circle $C_2AB$, $B_3$ and $C_3$ - point of intersection of $BB_1$ and $CC_1$ with the circumcircle $ABC$. Well known $CO = BO = OA = OI$ and $HC = CB_3$. Note $HCB_3 \sim IOB$ (because $\angle HCB_3 = 2\angle HCA = 2\angle OCB = \angle BOI$ and they are isosceles) $C_3CB_3 \sim C_2OB$ (because $\angle HCB_3 = \angle BOI$ and $\angle C_2BO = 90^{\circ} - \angle B + \angle A= \angle C_3B_3C$) Now, \[ \frac{IO}{CH}\ = \frac{OB}{CB_3}\ = \frac{C_2O}{CC_3} \]But $CC_3$ = $C_1H$, so \[ \frac{C_1H}{CH}\ = \frac{C_2O}{CO} \]It is means that $OH$ and $C_1C_2$ are parallel.