$F$ is image of $E$ under reflection across $CI$ followed by symmetry about the midpoint $M$ of $BC$ and similarly $G$ is image of $D$ under reflection across $BI$ followed by symmetry about $M.$ Since $D \mapsto E$ is a perspectivity between $AB,AC,$ then it follows that $\mathbb{H}:F \mapsto G$ is a homography clearly with no fixed points $\Longrightarrow$ it is induced by two superposed congruent pencils, or the sides of a rigid angle whirling around its vertex $P.$ Parallel from $I$ to $AB$ cut $AC$ at $L.$ $U$ is reflection of $L$ on $CI$ and $V$ is reflection of $U$ about $M,$ then $U,V$ are the limiting points of $\mathbb{H}$ $\Longrightarrow$ $P$ is on the perpendicular bisector $AI$ of $UV.$ When $DE \parallel BC,$ then $IECG$ and $IDBF$ become rhombi, thus by symmetry, we have $\angle FIG=\angle ILE=\angle IUV,$ which implies that $I \equiv P$ $\Longrightarrow$ $\angle FIG=\angle BAC=\text{const}$ for all $r.$

Nice Solution, Luis González! One can also solve this problem with similar triangles. Let $P$ and $Q$ be the orthogonal projections of $I$ on $BC$ and $AB$, respectively. Since $I$ is the incenter, we have $QB=BP=PC=a$, and because $BD=CG=b$, we get $DQ=GP=a-b$. So, the triangles $IDQ$ and $IGP$ are congruent, since $IP=IQ$ and $\angle GPI = \angle DQI = 90$. Thus, we have $\angle IDQ = \angle IGP$, or $\angle ADE = \angle IGF$. Analogously, $\angle AED = \angle IFG$ Hence, trinangles $ADE$ and $IGF$ are similar, which means that $\angle FIG = \angle EAD = \angle BAC$, which is a fixed angle, solving the problem.