Point Outside a Square

The point $\rm{E}$ is outside the square $\rm{ABCD}$ so that $\rm{ABE}$ is an equilateral triangle. What is the measure of angle $\rm{CED}$ in degrees?
Source: NCTM Teachers of Mathematics, April 2006

Solution

Triangle $\rm{BCE}$ is isosceles, the base angles are congruent
$\measuredangle 1=\measuredangle 2$
The interior angles of triangle $\rm{BCE}$ sum to $180^{\circ}$
$\measuredangle 1+\measuredangle 2+60^{\circ}+90^{\circ}=180^{\circ}$
$\measuredangle 1+\measuredangle 2=30^{\circ}$
$\measuredangle 2=30^{\circ}/2=15^{\circ}$
Similarly, $\measuredangle 3=15^{\circ}$
$\measuredangle {\rm{CED}}=60^{\circ}-\measuredangle 2-\measuredangle 3$
$=60^{\circ}-15^{\circ}-15^{\circ}$
$=30^{\circ}$