Example 1 In $\triangle A B C$, $\angle A B C=60^{\circ}$, $\angle A C B=20^{\circ}, M$ is a point on the bisector of $\angle A C B$, $\angle M B C=20^{\circ}$. Find the degree measure of $\angle M A B$.
Solution: As shown in Figure 1, let the angle bisector of $\angle M B A$ intersect $A C$ at $D$, and connect $D M$.

Obviously, $B M$ bisects $\angle D B C$, and $C M$ bisects $\angle D C B$, which means $M$ is the incenter of $\triangle D B C$. Therefore,
$\angle M O B=\angle M D C=60^{\circ}$.
Thus, $\angle A D H=60^{\circ}=\angle M D B$.
Hence, point $A$ and point $M$ are symmetric with respect to $B D$.
Therefore, $\angle M A B=90^{\circ}-\angle D B A=70^{\circ}$.
Here, the proof that “point $A$ and point $M$ are symmetric with respect to $B D$” is based on “angle, side, angle”.