In a triangle $$ABC$$, $$D$$, $$E$$ are points on $$\overline{AC}$$, $$\overline{BC}$$ respectively such that $$\overline{DE} \parallel \overline{AB}$$ and $$DE = 5$$. If $$AB = 8$$ and $$\overline{AE}$$ bisects $$\angle DEB$$, then the length of $$\overline{CE} =$$ ____________

Since $$DE \parallel AB$$, alternate angles give $$\angle DEA = \angle EAB$$. As $$AE$$ bisects $$\angle DEB$$, $$\angle DEA = \angle AEB$$, so $$\angle AEB = \angle EAB$$ and triangle $$ABE$$ is isosceles with $$BE = AB = 8$$. Triangles $$CDE$$ and $$CAB$$ are similar in the ratio $$\frac{DE}{AB} = \frac{5}{8}$$, so $$\frac{CE}{CE + 8} = \frac{5}{8}$$. This gives $$8\,CE = 5\,CE + 40$$, so $$CE = \frac{40}{3} \approx 13.33$$.