Example 5 In $\triangle A B C$, $\angle A B C=50^{\circ}$, $\angle A C B=30^{\circ}, Q$ is a point inside the triangle, $\angle Q C A=$ $\angle Q A B=20^{\circ}$. Find the degree measure of $\angle Q B C$.
Solution: As shown in Figure 5, let the perpendicular bisector of $BC$ intersect $BA$ and $AC$ at $D$ and $E$, respectively, and let $F$ be the foot of the perpendicular. Connect $QE$, $BE$, and $DC$.

Given $\angle ACD = 20^{\circ} = \angle ACQ$ and $\angle DAC = 80^{\circ} = \angle QAC$, we know that point $D$ is symmetric to point $Q$ with respect to $AC$. Therefore,
$\angle AEQ = \angle AED = \angle FEC = 60^{\circ}$.
Since $\angle BEF = \angle FEC = 60^{\circ}$, we have
$\angle AEB = 60^{\circ} = \angle AEQ$.
Thus, points $B$, $Q$, and $E$ are collinear.
Then $\angle QBC = \angle EBC = 30^{\circ}$.
Note that $AC$ is the external angle bisector of $\angle QAB$ in $\triangle AQB$ (this is not immediately obvious). On the extension of $BA$, take a point $D$ such that $DA = QA$, making point $D$ the symmetric point of $Q$ with respect to $AC$. By using the perpendicular bisector of $BC$, we appropriately "flip" $\angle ABC$ to the position of $\angle DCE$.