Given triangle $ ABC$. Point $ O$ is the center of the excircle touching the side $ BC$. Point $ O_1$ is the reflection of $ O$ in $ BC$. Determine angle $ A$ if $ O_1$ lies on the circumcircle of $ ABC$.
1. **Identify the given elements and their properties:**
   - Triangle \( ABC \).
   - Point \( O \) is the center of the excircle touching side \( BC \).
   - Point \( O_1 \) is the reflection of \( O \) in \( BC \).
   - \( O_1 \) lies on the circumcircle of \( \triangle ABC \).

2. **Determine the angles involving \( O \) and \( O_1 \):**
   - Since \( O \) is the center of the excircle opposite \( A \), it lies on the angle bisector of \( \angle BAC \).
   - The excircle touches \( BC \) at a point \( D \), and \( O \) is the midpoint of the arc \( BDC \) not containing \( A \).

3. **Reflect \( O \) over \( BC \) to get \( O_1 \):**
   - The reflection \( O_1 \) of \( O \) in \( BC \) implies that \( O_1 \) is also on the angle bisector of \( \angle BAC \).

4. **Use the cyclic property of \( O_1 \) on the circumcircle:**
   - Since \( O_1 \) lies on the circumcircle of \( \triangle ABC \), the quadrilateral \( O_1ABC \) is cyclic.
   - This implies that \( \angle BO_1C = \angle BAC \).

5. **Calculate the angles:**
   - \( \angle OBC = 90^\circ - \frac{B}{2} \) because \( O \) is the excenter.
   - \( \angle BO_1O = \frac{B}{2} \) because \( O_1 \) is the reflection of \( O \).
   - Similarly, \( \angle CO_1B = \frac{C}{2} \).

6. **Relate the angles in the cyclic quadrilateral:**
   - Since \( O_1ABC \) is cyclic, \( \angle BAC = \angle BO_1C \).
   - Therefore, \( \angle BO_1C = \frac{B}{2} + \frac{C}{2} \).

7. **Set up the equation:**
   - Given that \( \angle BAC = \angle BO_1C \), we have:
     \[
     \angle BAC = \frac{B}{2} + \frac{C}{2}
     \]
   - Using the angle sum property of a triangle, \( A + B + C = 180^\circ \), we can substitute \( B + C = 180^\circ - A \):
     \[
     \angle BAC = \frac{180^\circ - A}{2}
     \]

8. **Solve for \( \angle A \):**
   - Equate the expressions for \( \angle BAC \):
     \[
     A = \frac{180^\circ - A}{2}
     \]
   - Multiply both sides by 2:
     \[
     2A = 180^\circ - A
     \]
   - Add \( A \) to both sides:
     \[
     3A = 180^\circ
     \]
   - Divide by 3:
     \[
     A = 60^\circ
     \]

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