Given triangle $ABC$, let $D$, $E$, $F$ be the midpoints of $BC$, $AC$, $AB$ respectively and let $G$ be the centroid of the triangle. For each value of $\angle BAC$, how many non-similar triangles are there in which $AEGF$ is a cyclic quadrilateral?
1. **Identify the given elements and properties:**
   - Triangle \(ABC\) with midpoints \(D\), \(E\), and \(F\) of sides \(BC\), \(AC\), and \(AB\) respectively.
   - \(G\) is the centroid of the triangle.
   - \(AEGF\) is a cyclic quadrilateral.

2. **Use properties of midpoints and centroids:**
   - Since \(D\), \(E\), and \(F\) are midpoints, \(EF \parallel BC\), \(FD \parallel AC\), and \(DE \parallel AB\).
   - The centroid \(G\) divides each median in the ratio \(2:1\).

3. **Analyze the cyclic quadrilateral condition:**
   - For \(AEGF\) to be cyclic, the opposite angles must sum to \(180^\circ\). Thus, \(\angle AEG + \angle AFG = 180^\circ\).

4. **Use the properties of parallel lines and angles:**
   - Since \(EF \parallel BC\), \(\angle AFE = \angle ABC\).
   - Similarly, \(\angle AEG = \angle ACB\).

5. **Establish similarity of triangles:**
   - Since \(\angle AEG = \angle ACB\) and \(\angle AFE = \angle ABC\), we need to show that \(\triangle GBD \sim \triangle BAD\).
   - \(\angle GBD = \angle ABD\) and \(\angle GDB = \angle BDA\).

6. **Use the centroid property:**
   - The centroid \(G\) divides \(AD\) in the ratio \(2:1\), so \(AD = 3GD\).

7. **Set up the proportion:**
   - From \(\triangle GBD \sim \triangle BAD\), we have:
     \[
     \frac{BD}{AD} = \frac{GD}{BD}
     \]
   - This implies:
     \[
     BD^2 = AD \cdot GD = 3GD \cdot GD = 3GD^2
     \]
   - Therefore:
     \[
     BC^2 = 12GD^2 = \frac{12AD^2}{9} = \frac{4AD^2}{3}
     \]
   - Solving for \(AD\):
     \[
     AD = BC \sin 60^\circ
     \]

8. **Verify the cyclic condition:**
   - If \(AD = BC \sin 60^\circ\), then:
     \[
     \frac{BD}{AD} = \frac{GD}{BD}
     \]
   - This confirms \(\triangle GBD \sim \triangle BAD\), and thus \(\angle AGE = \angle DGB = \angle ABD = \angle AFE\).

9. **Determine the number of non-similar triangles:**
   - The locus of point \(A\) must satisfy both \(AD = BC \sin 60^\circ\) and a fixed \(\angle BAC\).
   - This results in two distinct points where these conditions are met.

Conclusion:
There are two non-similar triangles for each value of \(\angle BAC\) where \(AEGF\) is a cyclic quadrilateral.

The final answer is \(\boxed{2}\).