Given a segment $AB$ in the plane. Let $C$ be another point in the same plane,$H,I,G$ denote the orthocenter,incenter and centroid of triangle $ABC$. Find the locus of $M$ for which $A,B,H,I$ are concyclic.
1. **Understanding the Problem:**
   We are given a segment \( AB \) in the plane and a point \( C \) such that \( H \), \( I \), and \( G \) denote the orthocenter, incenter, and centroid of triangle \( ABC \), respectively. We need to find the locus of point \( M \) for which \( A, B, H, I \) are concyclic.

2. **Angles in Cyclic Quadrilateral:**
   For points \( A, B, H, I \) to be concyclic, the sum of opposite angles must be \( 180^\circ \). We know:
   \[
   \angle AHB = 180^\circ - \angle C
   \]
   and
   \[
   \angle AIB = 90^\circ + \frac{\angle C}{2}
   \]
   For \( A, B, H, I \) to be concyclic, these angles must be equal:
   \[
   180^\circ - \angle C = 90^\circ + \frac{\angle C}{2}
   \]

3. **Solving for \( \angle C \):**
   \[
   180^\circ - \angle C = 90^\circ + \frac{\angle C}{2}
   \]
   \[
   180^\circ - 90^\circ = \angle C + \frac{\angle C}{2}
   \]
   \[
   90^\circ = \frac{3\angle C}{2}
   \]
   \[
   \angle C = 60^\circ
   \]

4. **Constructing Points \( E \) and \( F \):**
   Let \( E \) and \( F \) be points on segment \( AB \) such that \( AE = EF = FB \). This means \( E \) and \( F \) divide \( AB \) into three equal parts.

5. **Midpoints and Parallel Lines:**
   Let \( M \) and \( N \) be the midpoints of \( AC \) and \( BC \), respectively. Since \( G \) is the centroid, it divides the medians in a 2:1 ratio. We have:
   \[
   \frac{BG}{GM} = 2 = \frac{BE}{EA}
   \]
   This implies \( GE \parallel AC \) and similarly \( GF \parallel BC \).

6. **Angle \( \angle EGF \):**
   Since \( \angle C = 60^\circ \), the angle between \( GE \) and \( GF \) is:
   \[
   \angle EGF = 60^\circ
   \]

7. **Locus of \( G \):**
   The locus of \( G \) is the circle passing through \( E \) and \( F \) that forms a \( 120^\circ \) arc. We only consider the part of the circle in the same half-plane as \( C \) with respect to \( AB \).

8. **Constructing Point \( C \):**
   Given any point \( G' \) on this arc, we can draw parallels through \( A \) and \( B \) to \( GE \) and \( GF \), respectively, to determine the position of point \( C \).

The final answer is the circle passing through \( \boxed{ E } \) and \( F \) forming a \( 120^\circ \) arc in the same half-plane as \( C \) with respect to \( AB \).