# 1951 AHSME Problems/Problem 46

## Problem $AB$ 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: $\textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies}$ $\textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C$

## Solution $[asy] pair O=(0,0), A=(-1,0), B=(1,0), C=(-0.5,0.5sqrt(3)), D=(-0.5,-0.5sqrt(3)), E=(0.5,-0.5sqrt(3)), P=(0,-1); draw(circle(O,1)); label("A",A,W); label("B",B,E); label("O",O,NE); label("P",P,S); label("C",C,NW); label("D",D,SW); label ("E",E,SE); dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); dot(P); draw(A--B); draw(C--D--E--cycle); draw(C--P,dashed); [/asy]$ 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 $\boxed{\textbf{(A)}\ \text{bisects the arc }AB}$.

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