One more

The problem is easier than I first thought. 

By the given information, we know that there are three right-angled triangles in the diagram. We know that $m\angle AEB=m\angle BEC=m\angle CED= 60^{\circ}$. We can use this information to determine that every right triangle is also a 30-60-90 triangle. We also know that $AE=24$.

A 30-60-90 triangle is a special kind of right triangle where the ratio of the side lengths are $1:\sqrt{3}:2$. $\overline{AE}$ is the longest side length because it is the hypotenuse of the largest right triangle in the diagram. $\overline{BE}$ is the shortest side length of $\triangle ABE$ because this side is opposite the smallest angle. We mentioned earlier what the ratio of the side lengths are, so we can determine the length of $\overline{BE}$ without doing anything too computationally demanding. 

Of course, the ultimate goal is to figure out the length of $\overline{CE}$. If you look at $\triangle BCE$, carefully, you will notice that we are in an identical situation to when we solved for $BE$. Notice that $\overline{BE}$ is the hypotenuse of this triangle, and $\overline{CE}$ is the shortest side length since it is opposite the 30º angle. We can use the same $1:\sqrt{3}:2$ relationship of the side lengths to find the missing length.