Given triangle $ABC$, the three  altitudes intersect at point $H$. Determine all points $X$ on the side $BC$ so that the symmetric of $H$ wrt  point $X$ lies on the circumcircle of triangle $ABC$.
1. **Understanding the Problem:**
   - We are given a triangle \(ABC\) with altitudes intersecting at the orthocenter \(H\).
   - We need to find all points \(X\) on the side \(BC\) such that the reflection of \(H\) with respect to \(X\) lies on the circumcircle of \(\triangle ABC\).

2. **Reflection and Circumcircle:**
   - Let \(H'\) be the reflection of \(H\) with respect to \(X\).
   - By definition of reflection, \(X\) is the midpoint of the segment \(HH'\).
   - We need \(H'\) to lie on the circumcircle of \(\triangle ABC\).

3. **Homothety and Nine-Point Circle:**
   - The nine-point circle (NPC) of \(\triangle ABC\) passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from the orthocenter to the vertices.
   - The homothety centered at \(H\) with a ratio of \(-2\) maps the nine-point circle to the circumcircle of \(\triangle ABC\).

4. **Finding Points \(X\):**
   - Since \(H\) is the homothetic center, the reflection of \(H\) with respect to any point on the nine-point circle will lie on the circumcircle.
   - Therefore, the points \(X\) on \(BC\) that satisfy the condition are the points where the nine-point circle intersects \(BC\).

5. **Conclusion:**
   - The set of points \(X\) that satisfy the problem are the points on \(BC\) that lie on the nine-point circle of \(\triangle ABC\).

The final answer is the points on \( \boxed{ BC } \) that lie on the nine-point circle of \(\triangle ABC\).