Let $ABC$ be a triangle, such that its incenter $I$ and circumcenter $U$ are distinct. For all points $X$ in the interior of the triangle let $d(X)$ be the sum of distances from $X$ to the three (possibly extended) sides of the triangle. Prove: If two distinct points $P,Q$ in the interior of the triangle $ABC$ satisfy $d(P)=d(Q)$, then $PQ$ is perpendicular to $UI$.