$18 \cdot 34$ As shown, in $\square A B C D$, the diagonals $A C$ and $B D$ intersect at $O$. Take a point $E$ on $B C$ such that $E C=\frac{1}{4} B C, D E$ intersects $A C$ at $F$. Then $A O: O F: F C$ equals
(A) $3: 2: 1$.
(B) $4: 3: 1$.
(C) $5: 3: 2$.
(D) $7: 4: 3$.
[Solution] From the given, we know $E C=\frac{1}{4} B C$, and $A D=B C$, so $A D: E C=4: 1$,

and $A F: F C=A D: E C=4: 1$.
Therefore, $A O: F C=\frac{A F+F C}{2}: F C=\frac{4+1}{2}: 1=5: 2$.
Also, $O F=O C-F C=A O-F C$,
$\therefore \quad A O: O F: F C=5:(5-2): 2=5: 3: 2$.
Hence, the answer is $(C)$.