Example 4 As shown in Figure $17-5$, in $\triangle A B C$, $\angle A C B=30^{\circ}, \angle A B C=$ $50^{\circ}, M$ is a point inside the triangle, $\angle M A C=40^{\circ}, \angle M C B=20^{\circ}$. Find the degree measure of $\angle M B C$.
(Mathematical Bulletin, Issue 9, 1999, Problem 1208)
Take a point $E$ on the extension of $C M$ such that $\angle E B C=60^{\circ}$, and take a point $D$ on the extension of $B E$ such that $\angle D C B=40^{\circ}$. Connect $D A, D M, D C$, and $E A$. It is easy to see that $C A \perp D B, B A \perp D C$, which means $A$ is the orthocenter of $\triangle B C D$, so $\angle A D B=30^{\circ}$.

In $\triangle C D E$, it is easy to see that $\angle E C D=20^{\circ}, \angle B D C=80^{\circ}$, so $\angle D E C=80^{\circ}=\angle E D C$. Since $E C = D C$, and $A C$ is the angle bisector of $\angle D C E$, $A C$ is the perpendicular bisector of $D E$, thus $A D=A E, \angle A E D=30^{\circ}$, $\angle A E M=50^{\circ}$. Also, $\angle A M E=\angle M A C+\angle M C A=50^{\circ}$, so $A M=A E=A D$, which means $A$ is the circumcenter of $\triangle D E M$, thus $\angle M D E=\frac{1}{2} \angle M A E=40^{\circ}$.

From $\angle B D C=80^{\circ}$, we know that $D M$ bisects $\angle B D C$. Since $C M$ bisects $\angle B C D$, $M$ is the incenter of $\triangle B C D$, so $\angle M B C=\frac{1}{2} \angle D B C=30^{\circ}$.