6. In $\triangle A B C$, $D$, $E$, and $F$ are the midpoints of $B C$, $A C$, and $A B$ respectively, and $G$ is the centroid. For each value of $\angle B A C$, how many non-similar $\triangle A B C$ are there such that quadrilateral $A E G F$ is cyclic?
6. From the fact that $A, F, G, E$ are concyclic, we get $C G \times C F=C E \times C A$, i.e., $\frac{2}{3} m_{c}^{2}=\frac{1}{2} b^{2}$, where the median $m_{c}$ satisfies $4 m_{c}^{2}+c^{2}=2\left(a^{2}+b^{2}\right)$. Therefore, $b^{2}+c^{2}=2 a^{2}$, and by the cosine rule, $b^{2}+c^{2}-2 b c \cdot \cos \alpha=a^{2}$, which leads to $b^{2}-4 b c \cos \alpha+c^{2}=0$. Let $\lambda=\frac{b}{c}$, then $\lambda^{2}-4 \lambda \cos \alpha+1=0$, where $\alpha=\angle B A C$.

When $4 \cos ^{2} \alpha \leqslant 1$ i.e., $|\cos \alpha| \leqslant \frac{1}{2}$, the above equation has no real solutions. When $|\cos \alpha|>\frac{1}{2}$ but $\cos \alpha > 60^{\circ}$, there are no triangles $\triangle A B C$ that meet the requirements.

When $\alpha \leqslant 60^{\circ}$, the above equation has real solutions $\lambda_{1}, \lambda_{2}$, and $\lambda_{1}, \lambda_{2}$ are reciprocals of each other. Therefore, for each $\alpha \leqslant 60^{\circ}$, there is exactly one non-similar triangle $\triangle A B C$ that meets the requirements.