17.51 In $\triangle A B C$, $A C: C B=3: 4, \angle C$'s exterior angle bisector intersects the extension of $B A$ at $P$ ($A$ is between $P$ and $B$), then $P A: A B$ is
(A) $1: 3$.
(B) $3: 4$.
(C) $4: 3$.
(D) $3: 1$.
(E) $7: 1$.
(12th American High School Mathematics Examination, 1961)
[Solution] Draw $P A^{\prime}$ such that $\angle C P A^{\prime}=\angle A P C$ intersecting the extension of $B C$ at $A^{\prime}$. Then $\triangle A P C \cong \triangle A^{\prime} P C$.
Thus, $A C=A^{\prime} C, P A=P A^{\prime}$.
In $\triangle P B A^{\prime}$, since $P C$ bisects $\angle A^{\prime} P B$, we have $\frac{B C}{C A^{\prime}}=\frac{P B}{P A^{\prime}}$, or $\frac{B C}{C A}=\frac{P B}{P A}=\frac{4}{3}$,
and $\frac{A B}{P A}=\frac{P B-P A}{P A}=\frac{P B}{P A}-1=\frac{4}{3}-1=\frac{1}{3}$, hence $P A: P B=3: 1$.
Therefore, the answer is $(D)$.