40. The angle $A$ at the vertex of the isosceles triangle $A B C$ is $100^{\circ}$. On the ray $A B$, a segment $A M$ is laid off, equal to the base $B C$. Find the measure of angle $B C M$.
83.40. Answer. $10^{\circ}$. Indeed, let's lay off on the ray $A M$ a segment $B N$, equal in length to $B C$. Then, in the isosceles triangle $B N C$, the line $C M$ is the bisector of angle $C$. This follows from the fact that $B M / B N = C B / C N$. Since $\angle B C N = 20^{\circ}$, then $\angle B C M = 10^{\circ}$.