Example 1 Find a point $P$ on the plane of equilateral $\triangle A B C$ such that $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ are all isosceles triangles. How many points $P$ with this property are there?
Solution: Let any side of the known triangle be denoted as $a$. When $a$ is the base of the sought isosceles triangle, point $P$ lies on the perpendicular bisector of $a$; when $a$ is one of the legs of the sought isosceles triangle, point $P$ lies on the circumference of the circle with the vertex of $\triangle ABC$ as the center and $a$ as the radius. Thus, the 3 perpendicular bisectors and 3 circles yield a total of 10 intersection points (as shown in Figure 1).

Therefore, there are 10 points $P$ that satisfy the conditions. These can be categorized into 4 types: the first type is the intersection of the 3 perpendicular bisectors $\left(P_{1}\right)$. The 2nd, 3rd, and 4th types are the intersections of the circles and the perpendicular bisectors, each with 3 points $\left(P_{2}, P_{3}\right.$, $\left.P_{4}\right),\left(P_{5}, P_{6}, P_{7}\right),\left(P_{8}, P_{9}, P_{10}\right)$.