SAMC

Let $\alpha =BC$, $\beta =CA$, $\gamma =AB$. Without loss of generality we may assume that $\beta \ge \gamma$. Let $M$ be the midpoint of $BC$, and let $A^{\prime }$, $I^{\prime }$, and $G^{\prime }$ be the orthogonal projections of $A$, $I$ and $G$ on $MB$.



We have $$M{I}^{\prime }=MB-B{I}^{\prime }=\frac{\alpha }{2}-(s-\beta )=\frac{\beta -\gamma }{2},$$ and because $${\gamma }^2-A^{\prime }{B}^2={\beta }^2-(\alpha -A^{\prime }B{)}^2,$$ we have $$\begin{array}{rl}& M{G}^{\prime }=\frac{1}{3}M{A}^{\prime }=\frac{1}{3}(MB-A^{\prime }B)\\ =& \frac{1}{3}\left(\frac{\alpha }{2}-\frac{{\alpha }^2-{\beta }^2+{\gamma }^2}{2\alpha }\right)=\frac{{\beta }^2-{\gamma }^2}{6\alpha }.\end{array}$$ We conclude that $IG\perp BC$ is equivalent to $I^{\prime }$ coincides to $G^{\prime }$, that is $M{I}^{\prime }=M{G}^{\prime }$. The last relation is equivalent to $$\frac{\beta -\gamma }{2}=\frac{{\beta }^2-{\gamma }^2}{6\alpha }$$ or $$(\beta -\gamma )(\beta +\gamma -3\alpha )=0$$ and the conclusion follows.