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 $Q$. It is known 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 $\triangle A B C$?
Solution: As shown in Figure 1, take point $D$ on the extension of $AB$ such that $BD = BP$; take point $E$ on the extension of $AQ$ such that $QE = QB$. Connect $PD$ and $PE$, then
$$
AD = AB + BP = AQ + QB = AE,
$$

and $\triangle ADP \cong \triangle AEP$.
Therefore, $\angle AEP = \angle ADP = \frac{1}{2} \angle ABC = \angle QBC$, which means $\angle QEP = \angle QBP$.
The following proof will use the following lemma.
Lemma In isosceles $\triangle ABC$, $AB = AC$, a point $P$ in the plane satisfies $\angle ABP = \angle ACP$, then point $P$ is on the perpendicular bisector of $BC$ or on the line segment $BC$. (Proof omitted)

In isosceles $\triangle QBE$, by the lemma, point $P$ is on the perpendicular bisector of $BE$ or on the line segment $BE$.
(1) If point $P$ is on the perpendicular bisector of $BE$, then $PB = PE$. Therefore,
$$
PD = PE = PB = BD.
$$

Thus, $\triangle PBD$ is an equilateral triangle.
Hence, $\angle ABC = 120^{\circ}$, which contradicts $\angle BAC = 60^{\circ}$.
(2) If point $P$ is on the line segment $BE$, then $C$ and $E$ coincide. It is easy to get $\angle ABC = 80^{\circ}, \angle ACB = 40^{\circ}$.
Combining (1) and (2), the possible values are
$$
\angle ABC = 80^{\circ}, \angle ACB = 40^{\circ}, \angle BAC = 60^{\circ}.
$$

From the reasoning process of this problem, a more general conclusion can be easily derived:

In $\triangle ABC$, $AP$ bisects $\angle BAC$, intersecting $BC$ at point $P$, $BQ$ bisects $\angle ABC$, intersecting $CA$ at point $Q$, $AB + BP = AQ + QB$. Let $\angle BAC = \alpha$.
(1) If $\alpha \in [60^{\circ}, 180^{\circ})$, then
$$
\angle ABC = 2 \angle ACB = \frac{2}{3}(180^{\circ} - \alpha);
$$
(2) If $\alpha \in (0^{\circ}, 60^{\circ})$, then
$$
\angle ABC = 2 \angle ACB = \frac{2}{3}(180^{\circ} - \alpha)
$$

or $\angle ABC = 120^{\circ}, \angle ACB = 60^{\circ} - \alpha$.