The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$
1. Let \( A \) be the vertex of the acute triangle \( ABC \). Given that \( A \) is equidistant from the circumcenter \( O \) and the orthocenter \( H \), we need to determine the measure of angle \( A \).

2. Recall that the circumcenter \( O \) is the center of the circumcircle of \( \triangle ABC \), and the orthocenter \( H \) is the point where the altitudes of \( \triangle ABC \) intersect.

3. Since \( A \) is equidistant from \( O \) and \( H \), we have \( AO = AH \).

4. The distance \( AO \) is the circumradius \( R \) of \( \triangle ABC \).

5. To find \( AH \), we use the fact that in any triangle, the distance from a vertex to the orthocenter can be expressed in terms of the sides and angles of the triangle. Specifically, for vertex \( A \), the distance \( AH \) is given by:
   \[
   AH = 2R \cos A
   \]
   where \( R \) is the circumradius and \( A \) is the angle at vertex \( A \).

6. Given \( AO = AH \), we substitute \( AO = R \) and \( AH = 2R \cos A \) into the equation:
   \[
   R = 2R \cos A
   \]

7. Simplifying the equation, we get:
   \[
   1 = 2 \cos A
   \]
   \[
   \cos A = \frac{1}{2}
   \]

8. The angle \( A \) for which \( \cos A = \frac{1}{2} \) is:
   \[
   A = 60^\circ
   \]

9. Therefore, the measure of angle \( A \) is \( 60^\circ \).

The final answer is \( \boxed{60^\circ} \).