$17 \cdot 49$ In $\triangle A B C$, $B D$ is a median, $C F$ intersects $B D$ at $E$, and $B E=E D$, point $F$ is on $A B$, if $B F=5$, then $B A$ equals
(A) 10.
(B) 12.
(C) 15.
(D) 20.
(E) None of the above.
(10th American High School Mathematics Examination, 1959)
[Solution] Take a point $G$ on $CE$ such that $FE = EG$, and connect $DG$. From $\triangle EFB \cong \triangle EGD$, we can prove that $FB \parallel DG$ and $FB = DG$.
In $\triangle AFC$, $DG$ is the midline, so we have
$$
\begin{aligned}
& AF = 2DG = 10. \\
\therefore \quad & AB = AF + FB = 15.
\end{aligned}
$$

Therefore, the answer is $(C)$.