7. In the plane of square $ABCD$, there is a point $P$ such that $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$ are all isosceles triangles. Then, the number of points $P$ with this property is
(A) 9;
(B) 17;
(C) 1;
(D) 5.
7. (A)

Since the base of the isosceles triangle, with one side of the square as its waist, has endpoints 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 in pairs to form 8 points that 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 twice, intersecting at one point. These two perpendicular bisectors intersect with the aforementioned four circles precisely at the aforementioned eight points. Therefore, a total of 9 different intersection points are formed.