5. 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 $\angle B A C^{\circ}=60^{\circ}$, and $A B+B P=A Q+B Q$. What are the possible values of the angles of $\triangle A B C$?
Solution 1 (Yang Xiao): Extend $AB$ to $B'$ such that $BB' = BP$, and take $C'$ on the ray $QC$ such that $QC' = QB$, as shown in Figure 4 or Figure 5.
If $C$ and $C'$ do not coincide, then connect $B'C'$, $B'P$, $C'P$, and $BC'$.
$$
\begin{array}{l}
\because AB + BP = AQ + QB, \\
\therefore AB + BB' = AQ + QC', \text{ i.e., } AB' = AC'. \\
\because \angle B'AC' = 60^\circ, \therefore \triangle AB'C' \text{ is an equilateral triangle.} \\
\because AP \text{ bisects } \angle B'AC', \\
\therefore AP \text{ is the axis of symmetry of } \triangle AB'C'. \\
\therefore \angle AB'P = \angle AC'P, B'P = C'P. \\
\because \angle ABC = \angle ABQ + \angle CBQ \\
= \angle BB'P + \angle BPB', \\
\angle ABQ = \angle CBQ, \angle BB'P = \angle BPB', \\
\therefore \angle QBC' = \angle BB'P = \angle QC'P = \alpha. \\
\because BQ = QC', \text{ we have } \angle QBC' = \angle QC'B = \beta. \\
\therefore \angle PBC' = \angle PC'B = \alpha - \beta \text{ or } \beta - \alpha. \\
\therefore BP = C'P = B'P = BB'.
\end{array}
$$

Then $\angle BB'P = 60^\circ$, $\angle ABC = 120^\circ$.
$\because \angle BAC = 60^\circ$, contradiction, $\therefore C, C'$ coincide.
Similarly, we can get $B'P = CP$, as shown in Figure 6.
$$
\begin{array}{l}
\text{Then } \angle PB'C = \angle PCB'. \\
\because \angle BPB' = \angle BB'P = \\
\angle PB'C + \angle PCB' = \gamma, \\
\therefore \angle PB'C = \frac{\gamma}{2}, \\
\gamma + \frac{\gamma}{2} = 60^\circ, \gamma = 40^\circ. \\
\angle ABC = 2\gamma = 80^\circ.
\end{array}
$$

Solution 2 (Liu Wenguang):
In $\triangle ABQ$, by the Law of Sines, we have
$$
\frac{AQ}{\sin \frac{B}{2}} = \frac{AB}{\sin \left(\frac{B}{2} + 60^\circ\right)} = \frac{BQ}{\sin 60^\circ}.
$$

Thus, $AQ = \frac{AB \cdot \sin \frac{B}{2}}{\sin \left(\frac{B}{2} + 60^\circ\right)}, BQ = \frac{AB \cdot \sin 60^\circ}{\sin \left(\frac{B}{2} + 60^\circ\right)}$.
Similarly, $BP = \frac{AB \cdot \sin 30^\circ}{\sin \left(30^\circ + B\right)}$.
Since $AB + BP = AQ + BQ$, $\angle B + \angle C = 120^\circ$, and $0^\circ < \angle B < 120^\circ$,
$$
\therefore 1 + \frac{\sin 30^\circ}{\sin \left(B + 30^\circ\right)}
= \frac{\sin \frac{B}{2}}{\sin \left(\frac{B}{2} + 60^\circ\right)} + \frac{\sin 60^\circ}{\sin \left(\frac{B}{2} + 60^\circ\right)}.
$$
Thus, $1 + \frac{\sin 30^\circ}{\sin \left(B + 30^\circ\right)} = \frac{\cos \left(\frac{B}{4} - 30^\circ\right)}{\cos \left(\frac{B}{4} + 30^\circ\right)}$,
$$
\therefore \frac{\sin 30^\circ}{\sin \left(B + 30^\circ\right)} = \frac{\sin \frac{B}{4}}{\cos \left(\frac{B}{4} + 30^\circ\right)}.
$$
$$
\therefore \sin 30^\circ \cdot \cos \left(\frac{B}{4} + 30^\circ\right) = \sin \frac{B}{4} \cdot \sin \left(B + 30^\circ\right)
= \frac{1}{2}\left[\cos \left(\frac{3B}{4} + 30^\circ\right) - \cos \left(\frac{5B}{4} + 30^\circ\right)\right].
$$
Then $\cos \left(\frac{B}{4} + 30^\circ\right) + \cos \left(\frac{5B}{4} + 30^\circ\right)$
$$
\begin{aligned}
& = \cos \left(\frac{3B}{4} + 30^\circ\right). \\
\therefore & 2 \cos \left(\frac{3B}{4} + 30^\circ\right) \cos \frac{B}{2} = \cos \left(\frac{3B}{4} + 30^\circ\right).
\end{aligned}
$$

When $\cos \left(\frac{3B}{4} + 30^\circ\right) = 0$, equation (1) holds. Then $\frac{3\angle B}{4} + 30^\circ = 90^\circ$, hence $\angle B = 80^\circ$.

When $\cos \left(\frac{3B}{4} + 30^\circ\right) \neq 0$, from equation (1), we know $\cos \frac{B}{2} = \frac{1}{2}$. Then $\angle B = 120^\circ$, which contradicts the problem statement.
In summary, the internal angles of $\triangle ABC$ are
$\angle BAC = 60^\circ, \angle ABC = 80^\circ, \angle ACB = 40^\circ$.