$16 \cdot 3$ On the plane of equilateral triangle $ABC$, find a point $P$ such that $\triangle PAB$, $\triangle PBC$, and $\triangle PAC$ are all isosceles triangles. The number of points with this property is
(A) 1.
(B) 4.
(C) 7.
(D) 9.
(E) 10.
(China Junior High School Mathematics Correspondence Competition, 1986)
[Solution] Construct the perpendicular bisectors $A D, B E, C F$ of the sides of $\triangle A B C$, intersecting the sides at $D, E, F$ respectively, and their intersection point $P$ is the required point.

Extend $B E$ and construct $A P^{\prime}=A B$, and $B P^{\prime \prime}=A B$, then $P^{\prime}$ and $P^{\prime \prime}$ are the required two points.
Extend $E B$ such that $B P^{\prime \prime \prime}=A B$, then $P^{\prime \prime \prime}$ is another required point;
Similarly, extend $A D$ and $D A$, and we can find three points that meet the requirements;

Extend $C F$ and $F C$, and we can also find three points that meet the requirements.
In total, there are $1+3 \cdot(1+2)=10$ such points. Therefore, the answer is $(E)$.