1. **Assume $AB < AC$: This assumption helps us to orient the triangle and the points correctly for the argument that follows.

2. Define points and properties**:
- Let $D$ be the antipode of $M$ with respect to the incircle $\omega$ of $\triangle ABC$. This means $D$ is the point on $\omega$ such that $MD$ is a diameter of $\omega$.
- By a homothety argument, $ADN$ are collinear.
- Let $T$ be the tangency point of $\omega$ with $\overline{AC}$.
- Suppose lines $ADN$ and $AM$ intersect $\omega$ again at points $P$ and $Q$, respectively.

3. Angle properties and similarity:
- Since $D$ is the antipode of $M$ with respect to $\omega$, $\angle DPM$ is a right angle.
- $\triangle AIT$ and $\triangle AMP$ are directly similar right triangles because they share the angle at $A$ and both have a right angle.
- It follows that $\triangle AIM \sim \triangle ATP$, so $\angle AMD = \angle TPD$.

4. Parallel lines and isosceles triangle:
- Since $D$ is the midpoint of arc $\widehat{QT}$ of $\omega$, $QT \parallel BC$.
- $\triangle QMT$ is isosceles with $\angle QMT = \angle C$.

5. Angle calculations:
- Compute $\angle AIM$:
$$\angle AIM = \angle AIB + \angle BIM = 180^\circ - \frac{B-C}{2}$$
- Since $\angle AMI = \frac{C}{2}$, we have:
$$\angle MAI = \frac{B}{2} - C$$
- Therefore:
$$\angle ANM = \angle ACN + \angle NAC = C + \left(\frac{B}{2} - C\right) = \frac{B}{2}$$
- This implies:
$$\angle ANM = \angle IBM$$

6. Similarity of triangles:
- $\triangle BMI$ and $\triangle NMD$ are similar right triangles because they share the angle at $M$ and both have a right angle.
- This implies:
$$\frac{MN}{BM} = \frac{DM}{IM} = 2$$

7. Final calculation:
- Since $DM = 2IM$, we have:
$$BC = MN + (BM + NC) = 2MN$$

Thus, we have shown that $BC = 2MN$.

$\blacksquare$

The final answer is $\boxed{ BC = 2MN }$