16. In $\triangle A B C$
$$
A B=8, B C=7 \text {, }
$$
$C A=6$. Also, as shown in the figure, extend side $B C$ to point $P$, such that $\triangle P A B$ is similar to $\triangle P C A$. The length of $P C$ is
$$
\text { (A) } 7 \text {; (B) } 8 \text {; (C) } 9 \text {; (D) } 10 \text {; (E) } 11 .
$$
$$
\frac{16}{\mathrm{C}}
$$
16. From the similarity of the two triangles, we get $\frac{P A}{P B}=\frac{P C}{P A}=\frac{C A}{A B}$. Therefore, $\frac{P A}{P C+7}=\frac{P C}{P A}=\frac{6}{8}$, which leads to two equations $6(P C+7)=8 P A$ and $6 P A=8 P C$, from which we can find $P C=9$.