Finding the measure of an angle 
Let E be a point outside of the square $ABCD$ such $\triangle ABE$ is an equilateral triangle.  What is the measure of $\angle CED$, in degrees?

I need help with this problem. I made a diagram beforehand to help me with this problem. From doing this, I find that $m(\angle CAE)$ is the same as $m(\angle DBE)$. I think these angles are $150^\circ$ because the angles of a square are each $90^\circ$, plus when we have point $E$ makes an equilateral triangle (which has three $60^\circ$ angles). So $60^\circ + 90^\circ = 150^\circ$. This is all I have so far. 
 A: Correction: I labeled points of square differently. But our solutions mirror one another.
The statement of the problem is below the image. 

Note that both $\angle EBC$ and $\angle EAD$ are $150^\circ$ ($90^\circ$ (measure of an angle of the square + $60^\circ$ (measure of each of three angles in an equilateral triangle)). Note also that $AE=AD=EB=BC$, since the lengths of all sides of square and triangle are equal (for triangle $ABE$ to be equilateral, both $EA$ and $EB$ must be the same length as $AB$.)
Since $AE=AD$, triangle $ADE$ is isosceles (and likewise, triangle $EBC$ is isosceles), and so the angles opposite the equal sides are equal. Since we have that $\angle EBC$ and $\angle EAD$ are $150^\circ$, $m(\angle ADE) + m(\angle AED) = m(\angle CEB) + m(\angle BCE) = 180 - 50 = 30^\circ$.  So, in particular, $m(\angle AED) = m(\angle CEB) = 15^\circ$. So we have that $m(\angle CED) = m(\angle AEB) - 2m(\angle AED) = 60 - 2(15) = 30^\circ$.
A: Draw the figure.  For definiteness, I would, like most mathematicians do, arrange the labels of the square counterclockwise around the square.  So put $A$ at top right, $B$ at top left, $C$ at bottom left, and $D$ at bottom right.
Then $E$ is somewhere above the line segment $AB$. (I am doing this to make sure that we are using the same diagram.  Yours was I think labelled differently; if so, please relabel.)
It is clear that $\angle EBC$ is $150^\circ$ ($90+60$).  So the two angles $BEC$ and $BCE$ add up to $180-150$, which is $30$.  But $BE=BC$, so the two small angles are equal.  It follows that each is $15^\circ$. 
Note: I did not see your diagram, but it looks as if you were probably very close to a solution. The only thing that you likely missed is the fact that what I called $\triangle EBC$ is isosceles. 
