$17 \cdot 119$ In isosceles $\triangle A B C$, the base $B C$ is $\frac{1}{4}$ of the perimeter. A line parallel to the two legs is drawn from any point $P$ on the base, intersecting the two legs at $E, F$. Then the ratio of the perimeter of quadrilateral $A E P F$ to the perimeter of $\triangle A B C$ is
(A) equal to $\frac{1}{4}$.
(B) equal to $\frac{2}{3}$.
(C) equal to $\frac{3}{4}$.
(D) cannot be determined.
(China Jiangxi Province Nanchang City Junior High School Mathematics Competition, 1990)
[Solution] Let the length of the waist be $l$, and the perimeter of $\triangle A B C$ be $L$. From the given information, we have $L=2 A B+B C=2 l+\frac{L}{4}$, then $L=2 l \cdot \frac{4}{3}$.
Also, since $A E P F$ is a parallelogram,
$\therefore \quad P E=B E, P F=C F$,
$\therefore$ the perimeter of $\square A E P F$ is $2 l$.
Thus, the ratio of the perimeter of quadrilateral $A E P F$ to the perimeter of $\triangle A B C$ is
$$
2 l:\left(2 l \cdot \frac{4}{3}\right)=3: 4 \text {. }
$$

Therefore, the answer is (C).