Problem 1622

1. Identify the incenter and the incircle:
Let $I$ be the incenter of $\triangle ABC$. The incenter is the point where the angle bisectors of the triangle intersect, and it is the center of the incircle of $\triangle ABC$. The radius of the incircle is denoted by $r$.

2. **Construct the circle with diameter $A'B'$:**
Let $\omega$ be the circle with diameter $A'B'$. Since $\angle A'C'B' = 90^\circ$, point $C'$ lies on the circle $\omega$ by the property of the circle that states an angle inscribed in a semicircle is a right angle.

3. **Consider the case where $\omega$ coincides with the incircle:**
If $\omega$ coincides with the incircle of $\triangle ABC$, then the diameter of $\omega$ is $2r$. In this case, the segment $A'B'$ is exactly $2r$, which is the diameter of the incircle.

4. **Consider the case where $\omega$ and the incircle are different:**
Assume $\omega$ and the incircle of $\triangle ABC$ are different. Since $\omega$ has points on the sides of $\triangle ABC$, it intersects at least one of the sides at two different points.

5. **Construct a tangential triangle to $\omega$:**
Construct a tangential triangle to $\omega$ whose sides are parallel to the sides of $\triangle ABC$. This tangential triangle is similar to $\triangle ABC$ with a similarity ratio greater than 1. This implies that the inradius of this tangential triangle is greater than the inradius of $\triangle ABC$.

6. Compare the inradii:
Since the inradius of the tangential triangle to $\omega$ is greater than the inradius of $\triangle ABC$, it follows that the diameter of $\omega$ (which is $A'B'$) is greater than the diameter of the incircle of $\triangle ABC$.

7. Conclusion:
Therefore, $A'B' > 2r$.

$$\boxed{A'B' > 2r}$$