Let $k$ be the incircle of triangle $ABC$. It touches $AB=c, BC=a, AC=b$ at $C_1, A_1, B_1$, respectively. Suppose that $KC_1$ is a diameter of the incircle. Let $C_1A_1$ intersect $KB_1$ at $N$ and $C_1B_1$ intersect $KA_1$ at $M$. Find the length of $MN$.
1. **Using Pascal's Theorem:**
   By applying Pascal's Theorem on the degenerate hexagon \((A_1A_1C_1B_1B_1K)\), we get that the points \(C\), \(N\), and \(M\) are collinear. This is because Pascal's Theorem states that for any hexagon inscribed in a conic, the intersections of the pairs of opposite sides are collinear.

2. **Properties of the Incircle:**
   Given that \(KC_1\) is a diameter of the incircle, it implies that \(\angle KC_1A_1 = \angle KC_1B_1 = 90^\circ\). This is because the angle subtended by a diameter in a circle is a right angle.

3. **Cyclic Quadrilateral:**
   Since \(\angle NB_1M\) and \(\angle NA_1M\) are right angles, quadrilateral \(NA_1B_1M\) is cyclic. This follows from the fact that if a quadrilateral has opposite angles that sum to \(180^\circ\), it is cyclic.

4. **Angles in the Cyclic Quadrilateral:**
   In the cyclic quadrilateral \(NA_1B_1M\), we have:
   \[
   \angle NMB_1 = \angle C_1A_1B_1 = \angle C_1B_1A = \angle CB_1M
   \]
   This implies that \(C\) is the midpoint of \(MN\).

5. **Length Calculation:**
   Since \(C\) is the midpoint of \(MN\), we have:
   \[
   MN = 2 \times MC = 2 \times CB_1
   \]
   Given that \(CB_1\) is the length from \(C\) to the point where the incircle touches \(BC\), we can use the fact that:
   \[
   CB_1 = \frac{a + b - c}{2}
   \]
   Therefore:
   \[
   MN = 2 \times \frac{a + b - c}{2} = a + b - c
   \]

The final answer is \(\boxed{a + b - c}\).