$17 \cdot 64 \triangle ABC$ has side $AB$ 8 inches long, line $DEF$ is parallel to $AB$, intersects $AC$ at $D$, and intersects $BC$ at $E$. Furthermore, the extension of $AE$ bisects $\angle FEC$, if $DE$ is 5 inches long, then the length of $CE$ (in inches) is
(A) $\frac{51}{4}$.
(B) 13 .
(C) $\frac{53}{4}$.
(D) $\frac{40}{3}$.
(E) $\frac{27}{2}$.
(19th American High School Mathematics Examination, 1968)
［Solution］Assume $D E / / A B$, then $\angle G E F=\angle E A B$, and $\angle G E F=\angle G E C=\angle A E B=\angle E A B$, thus $\triangle A E B$ is an isosceles triangle.
Furthermore, by $D E / / A B$, we have $\triangle C D E \backsim \triangle C A B$, so $\frac{E C}{B C}=\frac{D E}{A B}$, which means $\frac{E C}{8+E C}=\frac{5}{8}$,
solving this gives $E C=\frac{40}{3}$.
Therefore, the answer is $(D)$.