ABC is an acute angled triangle with circumcenter O and orthocentre H. If AO=AH, then find the angle A.

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Explanation

To solve the problem, we need to find the angle A in triangle ABC given that AO = AH, where O is the circumcenter and H is the orthocenter.

### Step-by-Step Solution:

1. **Understanding the Distances**:
- The distance from vertex A to the circumcenter O is denoted as OA, which is equal to the circumradius R of the triangle.
- The distance from vertex A to the orthocenter H is denoted as AH, which can be expressed in terms of the circumradius R and angle A. Specifically, we have:
\[ AH = 2R \cos A \]

2. **Setting Up the Equation**:
- Given that AO = AH, we can write:
\[ R = 2R \cos A \]

3. **Simplifying the Equation**:
- We can divide both sides of the equation by R (assuming R ≠ 0):
\[ 1 = 2 \cos A \]

4. **Solving for Cosine**:
- Rearranging the equation gives:
\[ \cos A = \frac{1}{2} \]

5. **Finding the Angle A**:
- The angle A for which the cosine is \( \frac{1}{2} \) is:
\[ A = \frac{\pi}{3} \text{ radians} \quad \text{or} \quad A = 60^\circ \]

### Conclusion:
Thus, the angle A in triangle ABC is:
\[ \boxed{60^\circ} \]