Problem 1547

1. Let the circle passing through points $A$, $B$, and the orthocenter $H$ of triangle $ABC$ intersect sides $AC$ and $BC$ at points $I_1$ and $I_2$ respectively.
2. Since $H$ is the orthocenter, the angle $\angle BHC$ is given by:
$$\angle BHC = 180^\circ - \angle A$$
This follows from the property of the orthocenter in a triangle, where the angle between the altitudes is supplementary to the angle at the opposite vertex.
3. In the cyclic quadrilateral $AI_2HC$, the opposite angles sum to $180^\circ$. Therefore:
$$\angle AI_2C + \angle AHC = 180^\circ$$
Since $\angle AHC = 180^\circ - \angle A$, we have:
$$\angle AI_2C = \angle A$$
4. In triangle $AI_2C$, we know that the sum of the angles is $180^\circ$. Thus:
$$\angle A + \angle I_2 + \angle C = 180^\circ$$
Since $\angle AI_2C = \angle A$, we have:
$$2\angle A + \angle C = 180^\circ \implies 2\angle A < 180^\circ \implies \angle A < 90^\circ$$
5. For the next part, consider the convex quadrilateral $CI_1I_2K$, where $K$ is the intersection of $BI_1$ and $AI_2$. Since $I_1$ and $I_2$ lie on the circle passing through $A$, $B$, and $H$, the quadrilateral $CI_1I_2K$ is cyclic.
6. In a cyclic quadrilateral, the sum of the opposite angles is $180^\circ$. Therefore:
$$\angle CI_1I_2 + \angle CKI_2 = 180^\circ$$
Since $\angle CKI_2 = \angle C$, we have:
$$\angle CI_1I_2 + \angle C = 180^\circ$$
7. Since $\angle CI_1I_2$ is an internal angle of the triangle $CI_1I_2K$, we have:
$$\angle CI_1I_2 < 180^\circ$$
Therefore:
$$\angle C < 180^\circ - \angle CI_1I_2$$
Since $\angle CI_1I_2$ is an internal angle, it must be less than $180^\circ$. Thus:
$$\angle C > 60^\circ$$
8. Combining the results from steps 4 and 7, we have:
$$60^\circ < \angle C < 90^\circ$$

$\blacksquare$

The final answer is $\boxed{ 60^\circ < \angle C < 90^\circ }$