$17 \cdot 63$ In $\triangle A B C$, $A B=8, B C=7, C A=6$. Also, as shown in the figure, extend side $B C$ to point $P$ so that $\triangle P A B$ is similar to $\triangle P C A$. The length of $P C$ is
(A) 7 .
(B) 8 .
(C) 9 .
(D) 10 .
(E) 11 .
(37th American High School Mathematics Examination, 1986)
[Solution] If $\triangle P A B \backsim \triangle P C A$, then we have
$$
\frac{P A}{P B}=\frac{P C}{P A}=\frac{C A}{A B}=\frac{6}{8}=\frac{3}{4} .
$$

When $P C=7$, $P A=\frac{4}{3} P C=\frac{28}{3}$, then
$$
P B=P C+C B=14 \text {. }
$$

At this time
$$
\frac{P A}{P B}=\frac{2}{3} \neq \frac{3}{4} .
$$

When $P C=8$, $P A=\frac{4}{3} P C=\frac{32}{3}$, then
$$
P B=P C+C B=15 \text {. }
$$

At this time
$$
\frac{P A}{P B}=\frac{32}{45} \neq \frac{3}{4} .
$$

When $P C=9$, $P A=\frac{4}{3} P C=12$, then $P B=P C+C B=16$.
At this time
$$
\frac{P A}{P B}=\frac{12}{16}=\frac{3}{4} \text {. }
$$

Therefore, the answer is $(C)$.