Let $\triangle ABC$ be a triangle with sides $\overline{AB} = c$, $\overline{a} = \overline{BC} = c + d$ and $\overline{b} = \overline{AC} = c + 2d$. Let $O$ be the center of the incircle and $O_1$ be the centroid, and $D, E$ be the feet of the side $BC$, respectively. Let $\overline{OD} = r$, and the area of triangle $\triangle ABC$ be $P = rs$. Since $s = \frac{c + d + c + c + 2d}{2} = \frac{3}{2}(c + d)$, we have $r = \frac{P}{s} = \frac{2P}{3(c + d)}$. Now $P_{O_1BC} = \frac{1}{3}P_{ABC} = \overline{O_1E} \cdot \frac{1}{2}(c + d)$, i.e. $\overline{O_1E} = \frac{2P}{3(c + d)} = r$.

Quadrilateral $DOO_1E$ is a parallelogram from where we get $OO_1 \parallel DE$, i.e. $OO_1 \parallel BC$.