In the triangle $ABC$ we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, and $\angle MCB=55^\circ$, $\angle NBC=80^\circ$. Find $\angle NMC$.

[i]St.Petersburg folklore[/i]
1. Given the angles in triangle \(ABC\):
   \[
   \angle ABC = 100^\circ, \quad \angle ACB = 65^\circ
   \]
   We can find \(\angle BAC\) using the fact that the sum of angles in a triangle is \(180^\circ\):
   \[
   \angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 100^\circ - 65^\circ = 15^\circ
   \]

2. Given the points \(M \in AB\) and \(N \in AC\) with the angles:
   \[
   \angle MCB = 55^\circ, \quad \angle NBC = 80^\circ
   \]

3. To find \(\angle CMB\), we use the fact that the sum of angles around point \(C\) is \(360^\circ\):
   \[
   \angle CMB = 180^\circ - (\angle MBC + \angle BCM)
   \]
   Since \(\angle MCB = 55^\circ\) and \(\angle ABC = 100^\circ\), we have:
   \[
   \angle MBC = \angle ABC - \angle MCB = 100^\circ - 55^\circ = 45^\circ
   \]
   Therefore:
   \[
   \angle CMB = 180^\circ - (45^\circ + 55^\circ) = 180^\circ - 100^\circ = 80^\circ
   \]

4. Next, we consider the reflection of \(AB \equiv MB\) in \(CM\). This reflection cuts \(BN\) at \(N'\). Since \(C\) is the M-excenter of \(\triangle MBN'\), we have:
   \[
   \angle BCN' = 90^\circ - \frac{1}{2} \angle N'MB
   \]
   Given \(\angle CMB = 80^\circ\), we find:
   \[
   \angle N'MB = 180^\circ - \angle CMB = 180^\circ - 80^\circ = 100^\circ
   \]
   Thus:
   \[
   \angle BCN' = 90^\circ - \frac{1}{2} \times 100^\circ = 90^\circ - 50^\circ = 40^\circ
   \]

5. Since \(\angle BCA = 65^\circ\), we have:
   \[
   \angle BCN' = \angle BCA = 65^\circ
   \]
   This implies that points \(N'\) and \(N\) are identical, i.e., \(N' \equiv N\).

6. Finally, we conclude that:
   \[
   \angle NMC = \angle CMB = 80^\circ
   \]

The final answer is \(\boxed{80^\circ}\).