G77 (42-5, Israel) In $\triangle A B C$, $A P$ bisects $\angle B A C$, intersecting $B C$ at $P$, $B Q$ bisects $\angle A B C$, intersecting $C A$ at $Q$. It is known that $\angle B A C=60^{\circ}, A B+B P=A Q+Q B$. What are the possible values of the angles of $\triangle A B C$?
Extend $A B$ to $E$, such that $B P = B E$.
Let $\angle A B C = \angle B, \angle A C B = \angle C$.
Now, find a point $D$ on $A C$ or its extension, such that $A L = A E$, then by the given conditions, $Q D = Q B$.
Since $\triangle A P D \cong \triangle A P E$, we have
$$
\angle A D P = \angle E = \frac{1}{2} \angle B \text{, }
$$

and $\angle Q D B = \frac{1}{2} \angle A Q B$
$$
\begin{array}{l}
= \frac{1}{2}(\angle Q B C + \angle C) \\
= \frac{1}{4} \angle B + \frac{1}{2} \angle C
\end{array}
$$

We will discuss the following cases.
(1) When $C$ coincides with $D$, we have $\angle B = 80^{\circ}, \angle C = 40^{\circ}$. It is easy to verify that this satisfies $A B + B P = A C = A Q + Q B$.
(2) When $D$ is between $C$ and $Q$, we have
$$
\begin{aligned}
\angle B D P & = \angle A D P - \angle A D B \\
& = \frac{1}{2} \angle B - \left(\frac{1}{4} \angle B + \frac{1}{2} \angle C\right) \\
& = \frac{1}{4} \angle B - \frac{1}{2} \angle C
\end{aligned}
$$

At this time, we also have
$$
\begin{aligned}
\angle D P C & = \angle A D P - \angle C \\
& = \frac{1}{2} \angle B - \angle C \\
& = 2 \angle B D P
\end{aligned}
$$

Thus, $B P = D P$.
Applying the Law of Sines to $\triangle A B P$ and $\triangle A D P$, we get $\sin \angle B = \sin \angle A D P$.
Since $A D = A B + B P > A B$, it is impossible for $\angle B = \angle A D P$. Therefore, $A, B, P, D$ are concyclic. Hence,
this means
$$
\begin{array}{l}
\angle D P C = 60^{\circ}, \\
\frac{1}{2} \angle B - \angle C = 60^{\circ},
\end{array}
$$

and
$$
\angle B + \angle C = 120^{\circ} \text{, }
$$

which gives $\angle C = 0^{\circ}$, which is impossible.
(3) When $D$ is on the extension of $A C$, similarly, we have $P D = B P$, so $A, B, P, D$ are concyclic, but $P$ is inside $\triangle A B D$, which is impossible.
In summary, the only solution is $\angle B = 80^{\circ}, \angle C = 40^{\circ}$.