AHSME 1968 Problem 18

Side $AB$ of triangle $ABC$ has length 8 inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC$, and $E$ is on segment $BC$. Line $AE$ extended bisects angle $FEC$. If $DE$ has length $5$ inches, then the length of $CE$, in inches, is: $\text{(A) } \frac{51}{4} \text{(B) } 13 \text{(C) } \frac{53}{4} \text{(D) } \frac{40}{3} \text{(E) } \frac{27}{2}$