Let $ABC$ be an acute-angled, not equilateral triangle, where vertex $A$ lies on the perpendicular bisector of the segment $HO$, joining the orthocentre $H$ to the circumcentre $O$. Determine all possible values for the measure of angle $A$.

(U.S.A. - 1989 IMO Shortlist)
1. **Identify the midpoint of \( OH \):**
   The midpoint of the segment \( OH \) (joining the orthocenter \( H \) and the circumcenter \( O \)) is the nine-point center \( N \). This is a well-known property in triangle geometry.

2. **Perpendicularity condition:**
   Since vertex \( A \) lies on the perpendicular bisector of \( OH \), it follows that \( AN \perp OH \).

3. **Congruence of triangles:**
   Consider the triangles \( \Delta AON \) and \( \Delta AHN \). Since \( N \) is the midpoint of \( OH \) and \( AN \perp OH \), we have:
   \[
   \angle OAN = \angle HAN = 90^\circ
   \]
   Additionally, \( ON = NH \) because \( N \) is the midpoint of \( OH \).

4. **Equality of segments:**
   Since \( \Delta AON \) and \( \Delta AHN \) share the side \( AN \) and have \( ON = NH \), by the Hypotenuse-Leg (HL) congruence theorem, the triangles are congruent:
   \[
   \Delta AON \cong \Delta AHN
   \]

5. **Equality of distances:**
   From the congruence, we get:
   \[
   AO = AH
   \]

6. **Relating \( AO \) and \( AH \):**
   Recall that \( AO = R \) (the circumradius) and \( AH = 2R \cos A \) (a known result in triangle geometry). Therefore, we have:
   \[
   R = 2R \cos A
   \]

7. **Solving for \( \cos A \):**
   Simplifying the equation:
   \[
   1 = 2 \cos A \implies \cos A = \frac{1}{2}
   \]

8. **Determining the angle \( A \):**
   The value \( \cos A = \frac{1}{2} \) corresponds to:
   \[
   A = 60^\circ
   \]

9. **Verification:**
   Since \( \triangle ABC \) is acute-angled and not equilateral, \( A = 60^\circ \) is a valid solution.

\(\blacksquare\)

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