Example 3 As shown in Figure 4, in a $4 \times 4$ grid square,
construct a grid point $\triangle A B C$
$(A B=\sqrt{5}, B C=$
$\sqrt{13}, C A=\sqrt{10}$ ). How many
grid point triangles congruent to
$\triangle A B C$ (including
$\triangle A B C)$ can be
constructed in Figure 4?
(1) As shown in Figure 5, $\triangle ABC$ is an inscribed lattice triangle in a $3 \times 3$ square, and there are 4 such $3 \times 3$ squares in Figure 4.
(2) As shown in Figure 5, one vertex of the triangle is at a vertex of the $3 \times 3$ square (point $C$, which has 4 possible positions), and the other two vertices $(A, B)$ are on the two adjacent sides of this vertex.

Solution 1: First, determine how many lattice $\triangle ABC$ can be formed in one $3 \times 3$ square. This can be done in two steps (Figure 6).
(1) First, find the side $AB$, whose endpoints are on the adjacent sides of the square (corresponding to 3 vertices of the square), with 2 possible choices, $AB$ and $A_1B_1$.
(2) Then, take the 4th vertex, which is not on the adjacent sides, as the 3rd vertex $C$ of the triangle, with only 1 possible choice.
Multiplying these, we get $2 \times 1 = 2$ lattice triangles.
Since a $3 \times 3$ square has 4 pairs of adjacent sides, there are a total of $2 \times 4 = 8$ lattice triangles.

Since there are 4 $3 \times 3$ squares in a $4 \times 4$ square, the number of lattice triangles congruent to $\triangle ABC$ is $8 \times 4 = 32$.

Analysis: From Solution 1, we know that once $AB$ is determined, $C$ is uniquely determined, so we only need to find the sides of length $\sqrt{5}$ in Figure 6.

Solution 2: In a $3 \times 3$ square, once $AB = \sqrt{5}$ is determined, $\triangle ABC$ is uniquely determined, so we only need to determine the number of inscribed squares with side length $\sqrt{5}$. Directly drawing the diagram yields 2 such squares (Figure 7).

Since there are 4 $3 \times 3$ squares in a $4 \times 4$ square, we get a total of 8 squares with side length $\sqrt{5}$. Each such square corresponds to 4 triangles, yielding a total of $4 \times 8 = 32$ triangles congruent to $\triangle ABC$.

Analysis: Here, a corresponding technique is used, converting the calculation of triangles to the calculation of squares with side length $\sqrt{5}$.
(To be continued)