$18 \cdot 70$ As shown in the figure, in the isosceles trapezoid $A B C D$, $A B$ // $D C, A B=998, D C=1001, A D=1999$, point $P$ is on the line segment $A D$, then the number of points $P$ that satisfy the condition $\angle B P C=90^{\circ}$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) An integer not less than 3.
[Solution] The midpoint $M$ of $AD$ subtends a $90^{\circ}$ angle with $BC$;
Suppose $N$ is on $AD$ and $AN: ND=998: 1001$. From the fact that $\triangle ABN$ and $\triangle DCN$ are both isosceles triangles, we deduce that $N$ subtends a $90^{\circ}$ angle with $BC$. Noting that the circle with $BC$ as its diameter intersects $AD$ at most at two points, we conclude that the number of points sought is 2.
Therefore, the answer is $(C)$.