AHSME 1961 Problem 31

In $\triangle ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ($A$ is between $P$ and $B$). The ratio $PA:AB$ is: $\textbf{(A)}\ 1:3 \textbf{(B)}\ 3:4 \textbf{(C)}\ 4:3 \textbf{(D)}\ 3:1 \textbf{(E)}\ 7:1$