7. In the plane of square $ABCD$, there is a point $P$ such that $\triangle PAB, \triangle PBC, \triangle PCD, \triangle PDA$ are all isosceles triangles. Then, the number of points $P$ with this property is:
(A) 9
(B) 17
(C) 1
(D) 5
7.【Analysis and Solution】Since the base endpoints of the isosceles triangle with one leg being the side of the square lie on the circle with the square's vertex as the center and the side length as the radius, such circles can be drawn four times, intersecting to form 8 points that all meet the requirements. The vertex of the isosceles triangle with the square's side as the base lies on the perpendicular bisector of the side, and such perpendicular bisectors can be drawn two times, intersecting at one point. These two perpendicular bisectors intersect with the aforementioned four circles at the aforementioned eight points, thus forming a total of 9 different intersection points.