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.