Arc of a Circle

$\overline{AB}$ is a diameter of a circle of radius $1$ unit. $\overline{CD}$ is a chord perpendicular to $\overline{AB}$ that cuts $\overline{AB}$ at $E$. If the arc $CAD$ is $2/3$ of the circumference of the circle, what is the length of the segment $\overline{AE}$?
Source: NCTM Mathematics Teacher, September 2006

Solution

Circumference = $2\pi(r)=2\pi(1)=2\pi$. Since measure of major arc $CAD=2/3(\mathrm{circumference})$, measure of minor arc $CBD=1/3(\mathrm{circumference})=2\pi/3$. Central angle $COD$ which tends arc $CBD$ therefore measures $2\pi/3$ radians or $120^\circ$. Triangle $COD$ is isosceles, hence altitude $OE$ is also the angle bisector of central angle $COD$. Triangle $OEC$ is a $30^\circ\!\textrm{-}60^\circ\!\textrm{-}90^\circ$ triangle with hypotenuse $OC$ equal $1$. Thus, side $\overline{OE}$ has length equal $1/2$.
$AE=AO+OE=1+1/2=3/2$ units

Answer: $3/2$ units

Alternative solution
Since chord $\overline{AB}$ is a diameter and chord $\overline{CD}$ is perpendicular to $\overline{AB}$, arc $AD$ is a mirror image of arc $AC$ in the $\overleftrightarrow{AB}$ line of reflection. Points $A,C$, and $D$ divide the circumference into three equal arcs each measuring $120^\circ$. Hence, $ACD$ is an equilateral triangle where the circumcenter $O$ is also the centroid of the triangle.
By the Concurrency of Medians of a Triangle theorem
$AO=(2/3)AE$
$AE=(3/2)AO=(3/2)(1)=3/2$