+981 \(\because \angle E = \angle E, \angle ABE = \angle CDE = 90º\therefore \text {By AA Similarity}, \triangle EBA \sim \triangle EDC.\\ \because \triangle EBA \sim \triangle EDC \therefore \frac{AB}{DC}=\frac{AE}{CE}.\\ \text{Filling in the numbers, we get}: \frac{4}{8}=\frac{5}{CE}.\\ \text{Solving for CE}: 4CE=5\cdot8\Rightarrow CE=10\)
The length of CE is 10.
I hope this helped,
Gavin.
