18.36 Given that $E$ is the midpoint of the diagonal $BD$ of square $ABCD$, take a point $F$ on $AD$ such that $DF=\frac{1}{3}DA$, and connect $EF$. Then the ratio of the area of $\triangle DEF$ to the area of quadrilateral $ABEF$ is
(A) $1: 2$.
(B) $1: 3$.
(C) $1: 5$.
(D) $1: 6$.
(E) $1: 7$.
(15th American High School Mathematics Examination, 1964)
[Solution] Given $D F=\frac{1}{3} D A$, then $S_{\triangle D F E}=\frac{1}{3} S_{\triangle D E A}$. Since $E$ is the midpoint of $B D$, then $S_{\triangle D E A}=\frac{1}{2} S_{\triangle D B A}$,
$$
\begin{aligned}
\therefore S_{\triangle D F E} & =\frac{1}{3} \cdot \frac{1}{2} S_{\triangle D B A} \\
& =\frac{1}{6} S_{\triangle D E A},
\end{aligned}
$$

thus $S_{\text {quadrilateral } A B E F}=\frac{5}{6} S_{\triangle D B A}$.
$\therefore \quad S_{\triangle D F E}: S_{\text {quadrilateral } A B E F}=1: 5$.
Therefore, the answer is $(C)$.