16.5 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.
(China High School Mathematics Competition, 1983)
[Solution 1] In a square, if an isosceles triangle is constructed with one side as the base, then because the vertex $P$ of this isosceles triangle must lie on the perpendicular bisector of this side, the opposite side must also be the base of another isosceles triangle with $P$ as the vertex. Therefore, we can discuss the following cases:
(1) In the square, all four sides are the bases of isosceles triangles.
In this case, the vertex $P$ must lie on the perpendicular bisectors of two pairs of opposite sides, so $P$ should be the center of the square, and there is only one such point, as shown in Figure (1).
[Solution 2] Point $P_{0}$ in the square clearly satisfies the condition.
Also, as shown in the right figure, $P_{1}, P_{2}$ also meet the requirements.
By symmetry, there are two more points on line $l$; there are 4 points on line $m$ (excluding $P_{0}$). In total, there are 9 points.
Therefore, the answer is $(A)$.