Example 4 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}$, $Q$ is a point inside the triangle, $\angle Q B A=$ $\angle Q C A=20^{\circ}$. Find the degree measure of $\angle Q A B$.
Solution: As shown in Figure 4, let $B Q$ intersect $A C$ at $D$, and draw a perpendicular from $D$ to $B C$ intersecting $Q C$ at $E$. Connect $B E$.

From $\angle Q 13 C=30^{\circ}=\angle A C B$, we know that $D E$ is the perpendicular bisector of $B C$.
From $\angle Q C B=10^{\circ}$, we know
$\angle E B C=10^{\circ}, \angle Q B E=20^{\circ}=\angle Q B A$.
From $\angle E D B=60^{\circ}=\angle E D C$, we know $\angle B D A$ $=60^{\circ}=\angle B D E$. Thus, point $A$ is symmetric to point $E$ with respect to $B D$.
$$
\begin{array}{l} 
\text { Then } \angle Q A B=\angle Q E B \\
=\angle E B C+\angle E C B=20^{\circ} .
\end{array}
$$

Here, we notice that $B Q$ is the angle bisector of $\angle A Q C$, so we consider taking a point $E$ on $Q C$ such that $\angle E B Q=\angle A B Q$, making point $E$ the symmetric point of $A$ with respect to $B Q$. To find such a point $E$, it is precisely the intersection of the perpendicular bisector of $B C$ with $Q C$. Furthermore, from $\angle Q B C=30^{\circ}=\angle A C B$, we think that the intersection point $D$ of $B Q$ and $A C$ should be another point on the perpendicular bisector of $B C$. Therefore, we chose the above method to find the symmetric point $E$ of $A$ with respect to $B Q$.