66. Given a triangle $A B C$, where $|A B|=|A C|$, $\widehat{B A C}=80^{\circ}$. Inside the triangle, a point $M$ is taken such that $\widehat{M} \overrightarrow{B C}=30^{\circ}, \widehat{M C B}=10^{\circ}$. Find $\widehat{A M C}$.
66. Let's draw (Fig. 12) the angle bisector of $\angle A$ and extend $B M$ until it intersects with the bisector at point N. Since $|B N|=|N C|$, then $\widehat{B N C}=120^{\circ} ;$ therefore, the angles $\widehat{B N A}, \widehat{C N A}$ are also $120^{\circ}$ each, $\widehat{N C A}=\widehat{N C M}=20^{\circ}, \quad$ i.e., $\triangle N M C=\triangle N C A,|M C|=$ $=|A C|, \quad \triangle A M C$ is isosceles.

Answer: $70^{\circ}$.