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Refer to the figure below.

Claim — The point $Q$ is the Miquel point of $BFEC$. Also, $\overline{QD}$ bisects $\angle BQC$. Proof. Inversion around the incircle maps line $EF$ to $(AIEF)$ and the nine-point circle of $△DEF$ to the circumcircle of $△ABC$ (as the midpoint of $EF$ maps to $A$, etc.). This implies $P$ maps to $Q$; that is, $Q$ coincides with the second intersection of $(AFIE)$ with $(ABC)$. This is the claimed Miquel point. The spiral similarity mentioned then gives $\frac{QB}{BF}=\frac{QC}{CE}$, so $\overline{QD}$ bisects $\angle BQC$. $\square$

Claim — We have $(QG;BC)=-1$, so in particular $\overline{GD}$ bisects $\angle BGC$. Proof. Note that $$-1=(AI;EF)\stackrel{Q}{=}(\overline{AQ}\cap \overline{EF},P;E,F)\stackrel{A}{=}(QG;BC).$$ The last statement follows from Apollonian circle, or more bluntly $\frac{GB}{GC}=\frac{QB}{QC}=\frac{BD}{DC}$. $\square$

Hence $\overline{QD}$ and $\overline{GD}$ are angle bisectors of $\angle BQC$ and $\angle BGC$. However, $\overline{QM}$ and $\overline{QG}$ are isogonal in $\angle BQC$ (as median and symmedian), and similarly for $\angle BGC$, as desired.