17・144 Given $A C \perp C E, D$ is the midpoint of $C E$, $B$ is the midpoint of $A C$, connect $A D, E B$, if $A D$ and $E B$ intersect at $F$, and $B C=C D=15$, then the area of $\triangle D E F$ is
(A) 50.
(B) $50 \sqrt{2}$.
(C) 75.
(D) $\frac{15}{2} \sqrt{105}$.
(E) 100.
(16th American High School Mathematics Examination, 1965)
[Solution] Draw $F G \perp C E$ at $G$, and connect $A E$. According to the problem, $F$ is the centroid of $\triangle A C E$.
$$
\begin{array}{l}
\therefore \quad F D=\frac{1}{3} A D, \\
\text { and } F G=\frac{1}{3} A C=\frac{1}{3} \cdot 30=10, \\
\therefore S_{\triangle D E F}=\frac{1}{2} \cdot D E \cdot F G=\frac{1}{2} \cdot 15 \cdot 10 \\
=75. \\
\end{array}
$$

Therefore, the answer is $(C)$.