17.65 In $\triangle A B C$, as shown in the figure, $\frac{C D}{D B}=\frac{3}{1}, \frac{A E}{E B}=\frac{3}{2}$, let $r=\frac{C P}{P E}$, where $P$ is the intersection of $C E$ and $A D$, then $r$ equals
(A) 3.
(B) $\frac{3}{2}$.
(C) 4.
(D) 5.
(E) $\frac{5}{2}$.
(14th American High School Mathematics Examination, 1963)
[Solution] Draw $D F / / A B$, by the property of parallel lines intercepting segments in proportion, we have $\frac{C F}{F E}=\frac{C D}{D B}=\frac{3}{1}, \quad$ and $\quad \frac{F D}{E B}=\frac{C D}{C B}=\frac{3}{4}$.
Thus, $C F=3 F E=3 F P+3 P E$,
and
$$
F D=\frac{3}{4} E B \text {. }
$$

Therefore,
$$
C P=C F+F P=4 F P+3 P E .
$$

Also, by $\triangle F D P \backsim \triangle E A P$, we have $\frac{F P}{P E}=\frac{F D}{A E}$, which means $F D=\frac{F P \cdot A E}{P E}$. But $A E=\frac{3}{2} E B$, so $F D=\frac{F P}{P E} \cdot \frac{3}{2} E B$.
$\therefore \quad \frac{3}{4} E B=\frac{3}{2} E B \cdot \frac{F P}{P E}$. Thus, $F P=\frac{1}{2} P E$,
then
$$
C P=4 \cdot \frac{1}{2} P E+3 P E=5 P E \text {. }
$$

Therefore, the answer is $(D)$.