In triangle $A B C$, a point $M$ is selected in its interior so that $\angle M A B=10^{\circ}$, $\angle M B A=20^{\circ}, \angle M C A=30^{\circ}$ and $\angle M A C=40^{\circ}$. Determine the value of $\angle M B C$.
Let $X$ be on side $B C$ so that $A M$ bisects $\angle A B X$. Let $Y$ be on side $A C$ so that $B M$ bisects $\angle A B Y$. Denote by $Z$ the intersection of these two lines; thus $M$ is the incenter of $\triangle A B Z$. Then, $\angle B M Z=90^{\circ} d g+\frac{1}{2} \angle B A Z=100^{\circ}$. On the other hand, $\angle B M C=180^{\circ}-\left(\angle B-20^{\circ}\right)-\left(\angle C-30^{\circ}\right)=100^{\circ}$ as well, so we conclude that $M, Z, C$ are collinear. Moreover, from $\angle A M B=150^{\circ}$ we obtain $\angle A Z B=120^{\circ}$, and hence the lines $A X, B Y, C M$ not only concur at $Z$ but form $60^{\circ}$ angles to each other.

Now, $\angle Z A C=\angle Z C A=30^{\circ}$. From this we deduce $Z A=Z C$, and hence line $B Z$ is the perpendicular bisector of side $A C$. Thus triangle $A B C$ is isosceles, with $A B=B C$. As $\angle A=50^{\circ}$, we get $\angle B=80^{\circ}$, and thus $\angle M B C=60^{\circ}$.