The orthocenter of triangle $ABC$ lies on its circumcircle. One of the angles of $ABC$ must equal:
(The orthocenter of a triangle is the point where all three altitudes intersect.)
$$
\mathrm a. ~ 30^\circ\qquad \mathrm b.~60^\circ\qquad \mathrm c. ~90^\circ \qquad \mathrm d. ~120^\circ \qquad \mathrm e. ~\text{It cannot be deduced from the given information.}
$$
1. Let \( \triangle ABC \) be a triangle with orthocenter \( H \) and circumcircle \( \Gamma \). Given that the orthocenter \( H \) lies on the circumcircle \( \Gamma \), we need to determine one of the angles of \( \triangle ABC \).

2. Recall that the orthocenter \( H \) is the point where the altitudes of the triangle intersect. The circumcircle \( \Gamma \) is the circle passing through all three vertices \( A, B, \) and \( C \) of the triangle.

3. Since \( H \) lies on \( \Gamma \), \( H \) must be one of the vertices of the triangle. This implies that \( \triangle ABC \) is a right triangle. In a right triangle, the orthocenter is the vertex where the right angle is located.

4. Let us assume without loss of generality that \( \angle C = 90^\circ \). Then, the orthocenter \( H \) coincides with vertex \( C \).

5. To confirm this, consider the properties of the angles in \( \triangle ABC \):
   - \( \angle A + \angle B + \angle C = 180^\circ \)
   - Since \( \angle C = 90^\circ \), we have \( \angle A + \angle B = 90^\circ \).

6. Therefore, one of the angles of \( \triangle ABC \) must be \( 90^\circ \).

The final answer is \( \boxed{ 90^\circ } \).