AoPS - Problem

Let $I$ be the incenter of $\triangle ABC$. Construct the circle $\omega$ with diameter $A'B'$. Obviously $C'$ lies on this circle. If $\omega$ coincides with the incircle of $\triangle ABC$, then the assumption is obvious. Now assume that $\omega$ and incricle of $\triangle ABC$ are different. Since $\omega$ has points on the sides of triangle $ABC$, it intersect at least one of the sides at two different points. Let's construct a tangential triangle to $\omega$ sides of which are parallel to sides of $ABC$. Then this triangle is similar to $\triangle ABC$ with coefficient of greater than 1. This means, that the inradius of this triangle $\omega$ is greater than the inradius of $\triangle ABC$. From this $A'B'>2r$ follows.