11.1. [7-8.6 (20 points)] On the bisector of angle $B A C$ of triangle $A B C$, a point $M$ is marked, and on the extension of side $A B$ beyond point $A$, a point $N$ is marked such that $A C=A M=1$ and $\angle A N M=\angle C N M$. Find the length of segment $A N$.
Answer: 1.

Solution. Let's place point $K$ on the extension of segment $NC$ beyond point $C$. Point $M$ lies on the bisectors of angles $BAC$ and $ANC$, which means it is equidistant from lines $AC$ and $NC$, and therefore lies on the bisector of angle $ACK$. Then, considering the equality of angles $M$ and $C$ in the isosceles triangle $AMC$, from the equality of alternate interior angles $CMA$ and $MCK$, we get that lines $AM$ and $NC$ are parallel. From this, it follows that alternate interior angles $MAC$ and $ACN$ are equal to $\angle BAC / 2$, and thus $\angle ANC = \angle BAC - \angle ACN = \angle BAC / 2$. We have obtained that triangle $NAC$ is isosceles and $AN = AC = AM = 1$.