In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$
is similar to $\triangle PCA$. The length of $PC$ is

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$
Since we are given that $\triangle{PAB}\sim\triangle{PCA}$, we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$.

Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$.

We also have $\frac{PA}{PC+7}=\frac{3}{4}$. Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$

Which we can further simplify to $\frac{16PC}{3}=3PC+21$

$\frac{7PC}{3}=21$

$PC=9\implies\boxed{\textbf{(C)}}$