$AB$ is a fixed diameter of a circle whose center is $O$. From $C$, any point on the circle, a chord $CD$ is drawn perpendicular to $AB$. Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always: 
$\textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies}$
$\textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C$
Draw in diameter $CE$. Note that, as it is inscribed in a semicircle, $\bigtriangleup CDE$ always has a right angle at $D$. Since it is given that $AB \perp CD$, $AB \parallel DE$ by congruent corresponding angles. $CP$ bisects $DE$, as well as the arc subtended by $DE$, $\widehat{DPE}$. Because $AB \parallel DE$, $CP$ always $\boxed{\textbf{(A)}\ \text{bisects the arc }AB}$.