1968 AMC 12 Problem 18

Put $G$ on the extension of $AE$ beyond $E$, so the bisector condition is $\angle FEG=\angle GEC$

After showing $BE=AB,$ use similarity of $CDE$ and $CAB$

Put $G$ on the extension of $AE$ beyond $E.$ The bisector condition gives $\angle FEG=\angle GEC.$ Since $FE\parallel AB,$ we have $\angle FEG=\angle BAE.$ Also, rays $EG$ and $EA$ are opposite, as are rays $EC$ and $EB,$ so $\angle GEC=\angle AEB.$ Thus triangle $ABE$ is isosceles and $BE=AB=8.$ Similar triangles $CDE$ and $CAB$ give $\frac{CE}{CE+8}=\frac58.$ Hence $8CE=5CE+40,$ so $CE=\frac{40}{3}.$

Therefore, the correct answer is D.