17.9 (IMO 42 Pre) In $\triangle A B C$, $A P$ bisects $\angle B A C$, intersecting $B C$ at $P$, and $B Q$ bisects $\angle A B C$, intersecting $C A$ at $Q$. Given that $\angle B A C=60^{\circ}$ and $A B+B P=A Q+Q B$, what are the possible values of the angles of $\triangle A B C$?
17.9 Let $A=\angle BAC, B=\angle ABC, C=\angle ACB$. On the extension of $AQ$, construct $E$ such that $QB=QE$. Connect $BE$, and let $BE \cap AP = F$. On the extension of $AB$, construct $D$ such that $BD=BP$. Then $\angle BDP = \frac{1}{2} \angle ABC = \angle ABQ$, so $BQ \parallel DP$, and $AD=AE$. Consider two cases: (1) $C=E$, then $BQ=QC$, so $\angle QBC = \angle QCB$.

Thus, $B = 2 \angle QBC = 2C$. Since $B + C = 120^\circ$, we have $B = 80^\circ, C = 40^\circ$. Verification shows that $B = 80^\circ, C = 40^\circ$, and $A = 60^\circ$ is a solution.
(2) $C \neq E$, if from the conditions we have $AD = AE$, and $PA$ bisects $\angle DAE$, hence $D$ and $E$ are symmetric with respect to line $AP$. From the angle relationships, we deduce that points $B, D, F, P$ are concyclic, leading to $\frac{1}{2}\left(\frac{B}{2} + C\right) = 30^\circ + C$, and $B + C = 120^\circ$, yielding $C = 0^\circ$, a contradiction. In this case, the triangle does not exist.