In $\triangle A B C$, $A P$ bisects $\angle B A C$, and $A P$ intersects $B C$ at $P$. $B Q$ bisects $\angle A B C$, and $B Q$ intersects $C A$ at $Q$. Given that $\angle B A C=30^{\circ}$ and $A B+B P=A Q+$ $Q B$. Find the possible values of the interior angles of $\triangle A B C$.

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Given $\triangle A B C$ with side lengths $a, b, c$.
Considering $S_{\triangle A B C}=S_{\triangle A E Q}+S_{\triangle B Q C}$. Therefore, $\frac{1}{2} a c \sin B=\frac{1}{2} c \cdot B Q \sin \frac{B}{2} + B Q \cdot a \sin \frac{B}{2}$.
Thus, $B Q=\frac{a c \sin B}{(a+c) \sin \frac{B}{2}}=\frac{2 a c}{a+c} \cos \frac{B}{2}$.
Since $B Q$ bisects $\angle A B C$, we have $\frac{A Q}{Q C}=\frac{c}{a}$, so $\frac{A Q}{A Q+Q C}=\frac{c}{a+c}$.
Therefore, $A Q=\frac{b c}{a+c}$, similarly $B P=\frac{a c}{b+c}$.
By the cosine rule, $b^{2}=a^{2}+c^{2}-2 a c \cos B$.
Thus, $(a+c)^{2}-b^{2}=4 \cos ^{2} \frac{B}{2} \cdot a c$.
Therefore, $A B+B P=A Q+Q B$.
$\Leftrightarrow c+\frac{d c}{b+c}=\frac{b c}{a+c}+\frac{2 a c}{a+c} \cos \frac{B}{2}$
$\Leftrightarrow \cos \frac{B}{2}=\frac{(a+c)^{2}-b^{2}+a b}{2 a(b+c)}=\frac{4 a c \cos ^{2} \frac{B}{2}+a b}{2 a(b+c)}$
$4 c \cos ^{2} \frac{B}{2}-2(b+c) \cos \frac{B}{2}+b=0$
Solving this, we get $\cos \frac{B}{2}=\frac{b}{2 c}$ or $\cos \frac{B}{2}=\frac{1}{2}$, so $\cos \frac{B}{2}=\frac{\sin B}{2 \sin C}$ or $B=120^{\circ}$.
Thus, $\sin \frac{B}{2}=\sin C$ or $B=120^{\circ}$
Therefore, $B=2 C$ or $B=120^{\circ}$.
1. When $B=2 C$, $A=30^{\circ}, B=100^{\circ}, C=50^{\circ}$
2. When $B=120^{\circ}$, $A=30^{\circ}, B=120^{\circ}, C=30^{\circ}$.