$16 \cdot 4$ In a right-angled $\triangle ABC$, where $AC < AB < BC$, in the plane of $\triangle ABC$, the number of points $P$ such that $\triangle PAB$ and $\triangle PBC$ are both isosceles triangles is
(A) 6.
(B) 9.
(C) 13.
(D) 14.
(China Hubei Province Huanggang Junior High School Mathematics Competition, 1993)
[Solution] First, the intersection point $P_{1}$ of the perpendicular bisectors $l, m$ of $B C$ and $A B$ is the desired point.

The four intersection points $P_{2}, P_{3}, P_{4}, P_{5}$ of the arcs drawn with $B, A$ as centers and $A B$ as radius with $l$ are also the desired points.

Similarly, four intersection points can be found on $m$ (the arcs drawn with $B, C$ as centers and $B C$ as radius intersecting with $m$).

Also, the circle with $A$ as center and $A B$ as radius intersects with the circle with $B$ as center and $B C$ as radius at two points $P_{6}, P_{7}$;

Similarly, the circle with $B$ as center and $A B$ as radius intersects with the circle with $C$ as center and $B C$ as radius at two points;

The circle with $A$ as center and $A B$ as radius intersects with the circle with $C$ as center and $B C$ as radius at one point (the other point coincides with $B$).
In total, there are 14 points that are the desired points.
Therefore, the answer is $(D)$.