$20 \cdot 10 A B$ is a fixed diameter of circle $O$, and a chord $C D$ is drawn from any point $C$ on the circle perpendicular to $A B$. When $C$ moves on the semicircle, the bisector of $\angle O C D$ intersects the circumference at point $P$, which is always
(A) bisects $A B$.
(B) trisects $\overparen{A B}$.
(C) varies.
(D) equidistant from $D$ and $A B$.
(E) equidistant from $B$ and $C$.
(2nd American High School Mathematics Examination, 1951)
[Solution] Extend $O O$ to intersect the circle at $E$, and connect $D E$, then
$$
\angle C D E=90^{\circ} .
$$

Also, $C D \perp A B$, so $A B / / D E$.
Since $C P$ is the angle bisector of $\angle O C D$, point $P$ is the midpoint of $D E$.
Given $D E / / A B$, point $P$ is also the midpoint of $A B$. Therefore, the answer is $(A)$.