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 $\angle B A C=60^{\circ}$ and $A B+B P=A Q+Q B$. What are the possible values of the angles of $\triangle A B C$?
Solution: Let $A=\angle BAC, B=\angle ABC, C=\angle ACB$.
On the extension of $AQ$, construct $E$ such that $QB=QE$, and connect $BE$ to intersect line $AP$ at $F$. On the extension of $AB$, construct $D$ such that $BD=BP$. Then,
$$
\angle BDP=\frac{1}{2} \angle ABC=\angle ABQ.
$$

Therefore, $BQ \parallel DDP$.
Thus, $AB+BP=AQ+QP$ is equivalent to
$AD=AE$.
Next, we discuss the cases where $E$ coincides with $C$ and where it does not.
(i) $C=E$. We have $BQ=QC$. Therefore,
$\angle QBC=\angle QCB$.
Thus, $B=2 \angle QBC=2 \angle QCB=2C$.
Since $B+C=180^{\circ}-A=120^{\circ}$,
we have $B=80^{\circ}, C=40^{\circ}$,
When $B=80^{\circ}, C=40^{\circ}$,
$\angle BDP=\frac{1}{2} \angle ABC=40^{\circ}=\angle ACP$,
$\angle DAP=\angle PAC, AP=PA$,
Therefore, $\triangle APD \cong \triangle APC$,
Hence, $AD=AC$.
From (1), we know that $AB+BP=AQ+QB$.
Thus, $B=80^{\circ}, C=40^{\circ}, A=60^{\circ}$ is a solution.
(ii) $C \neq E$. If $AB+BP=AQ+QB$, then $AD=AE$,
Since $PA$ bisects $\angle DAE$,
Thus, $D$ and $E$ are symmetric with respect to line $AP$. Therefore,
$\angle ADF=\angle AEB=\angle QBE$.
And $\angle BDP=\angle ABQ=\angle QBC$,
Thus, $\angle PDF=\angle PBF$.
Therefore, points $B, D, F, P$ are concyclic, hence,
$\angle BDF=\angle BPA$.
From (2), we have
$$
\begin{aligned}
\angle BDF & =\angle AEF \\
& =\frac{1}{2} \angle AQB=\frac{1}{2}\left(\frac{B}{2}+C\right). \\
\angle APB & =\frac{A}{2}+C=30^{\circ}+C
\end{aligned}
$$

From (4), we get
$$
\frac{1}{2}\left(\frac{B}{2}+C\right)=30^{\circ}+C.
$$

Thus, $\frac{B}{4}=30^{\circ}+\frac{C}{2}$.
Since $B+C=120^{\circ}$,
we have $30^{\circ}-\frac{C'}{4}=30^{\circ}+\frac{C}{2}$.
Thus, $C=0^{\circ}$, which is a contradiction.
Therefore, $\triangle ABC$ does not exist in this case.
Combining (i) and (ii), the possible values are
$$
B=80^{\circ}, C=40^{\circ}, A=60^{\circ}.
$$