Example 3 If 5 vertices of a regular nonagon are painted red, what is the minimum number of pairs of congruent triangles, all of whose vertices are red points?
(1992 Tianjin Team Test Question)
Solving: There are $C_{5}^{3}=10$ triangles with 5 red points as vertices, which we call red triangles. Let the circumference of the circumcircle of a regular nonagon be 9, and use the length of the arc opposite the chord to represent the chord length. In a regular nonagon, there are only 7 types of non-congruent triangles with its 9 vertices as vertices, with the lengths of their 3 sides being $(1,1,7),(1,2,6),(1,3,5)$, $(1,4,4),(2,2,5),(2,3,4),(3,3,3)$. Therefore, by the pigeonhole principle, there are at least 3 pairs of congruent red triangles. If there are at least 3 that are pairwise congruent, then the number of congruent triangle pairs is at least 4, so we assume below that these 3 pairs of congruent triangles are not congruent to each other.

Suppose two red triangles $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ are congruent. Because there are only 5 red vertices, these two red triangles must have at least 1 common point and at most 2 common points.
(1) Suppose two red triangles $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ have exactly one common vertex, without loss of generality, let $A_{1}$ and $B_{1}$ coincide (denoted as $A$). Since both are inscribed in the same circle, they must be symmetric about the diameter through $A$, at this time $A_{2}, A_{3}$, $B_{2}, B_{3}$ are the 4 vertices of an isosceles trapezoid $\left(A_{2} B_{2} / / A_{3} B_{3}, A_{2} A_{3}=B_{2} B_{3}\right)$, it is easy to see that there are at least 4 pairs of congruent red triangles: $\triangle A A_{2} A_{3}$ and $\triangle A B_{2} B_{3} ; \triangle A A_{2} B_{3}$ and $\triangle A B_{2} A_{3} ; \triangle A_{2} A_{3} B_{2}$ and $\triangle B_{2} B_{3} A_{2} ; \triangle A_{2} A_{3} B_{3}$ and $\triangle B_{2} B_{3} A_{3}$.
(2) Suppose two red triangles $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ share a common side, without loss of generality, let $A_{2}$ and $B_{2}$ coincide (denoted as $C$), $A_{3}$ and $B_{3}$ coincide (denoted as $D$), at this time $A_{1} B_{1} / / C D$. Thus, $A_{1} B_{1} D C$ is an isosceles trapezoid, yielding two pairs of congruent red triangles. Since there are at least 3 pairs of red triangles, there must be another pair of red triangles. Starting from this pair of congruent triangles (whether they have exactly one common vertex or exactly one common side), another pair of congruent red triangles can be found, hence there are at least 4 pairs of red triangles.

On the other hand, when 5 vertices $A_{1}, A_{4}, A_{5}, A_{6}, A_{7}$ of the regular nonagon $A_{1} A_{2} \cdots A_{9}$ are colored red, there are exactly 4 pairs of red triangles: $\triangle A_{1} A_{4} A_{5}$ and $\triangle A_{1} A_{7} A_{6} ; \triangle A_{1} A_{4} A_{6}$ and $\triangle A_{1} A_{7} A_{5} ; \triangle A_{4} A_{5} A_{6}$ and $\triangle A_{7} A_{6} A_{5} ; \triangle A_{4} A_{5} A_{7}$ and $\triangle A_{7} A_{6} A_{4}$.
In summary, the minimum number of pairs of congruent triangles is 4.