1. In $\triangle A B C$, $D, E, F$ are the midpoints of $B C$, $C A, A B$ 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 $A E G F$ is a cyclic quadrilateral?
1. From $A, E, G, F$ being concyclic, we get
$$
\begin{aligned}
\angle C G_{E} & =\angle B A C \\
= & \angle C E D .
\end{aligned}
$$

Let $C G$ intersect $D E$ at $M$. Then from (1),
it is easy to derive
$$
C M \times C G=C E^{2},
$$

i.e., $\frac{1}{2} m_{\mathrm{e}} \times \frac{2}{3} m_{\mathrm{c}}=\left(\frac{1}{2} b\right)^{2}$.
Here, the median
$$
m_{c}^{2}=\frac{1}{2} a^{2}+\frac{1}{2} b^{2}-\frac{1}{4} c^{2} .
$$

From (2) and (3), we get
$$
2 a^{2}=b^{2}+c^{2} \text {. }
$$

Thus, by the cosine rule and (4), we have
$$
b^{2}+c^{2}=4 b c \cos \angle B A C \text {. }
$$

Let $\frac{b}{c}=\lambda$, then
$$
\lambda^{2}+1=4 \lambda \cos \angle B A C .
$$
In (6), there are real solutions if and only if $\angle B A C \leqslant 60^{\circ}$. The solutions $\lambda_{1}, \lambda_{2}$ are reciprocals of each other, thus generating similar triangles. Therefore, for $\angle B A C \leqslant 60^{\circ}$, there is only one non-similar $\triangle A B C$ that satisfies the conditions. When $\angle B A C > 60^{\circ}$, there are no $\triangle A B C$ that satisfy the conditions.