At first 272, from the eight points consisting of the vertices and midpoints of the sides of a square, how many isosceles triangles can be formed by selecting three points?

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Considering that there are no equilateral triangles that meet the requirements in this problem, we classify the isosceles triangles by their vertices.
(1) Using the vertices of the square as the vertices of the isosceles triangle. As shown in Figure 3, if $A$ is the vertex,
then $\triangle A G H$, $\triangle A B D$, and $\triangle A E F$ all meet the requirements.
The same applies to points $B$, $C$, and $D$.
Therefore, there are
$$
3 \times 4=12 \text{ such triangles. }
$$
(2) Using the midpoints of the square's sides as the vertices of the isosceles triangle.

As shown in Figure 3, if $G$ is the vertex, then $\triangle G H F$ and $\triangle G B C$ both meet the requirements.
The same applies to points $H$, $E$, and $F$.
Therefore, there are $2 \times 4=8$ such triangles.
In summary, there are 20 triangles that meet the requirements.