Let $I_B$ and $I_C$ be the excenters opposite to $B$ and $C$, Let $N$ be the midpoint of $I_BI_C$, it suffices to show $\measuredangle IGM=\measuredangle MID$, or equivalently $\measuredangle IMB= \measuredangle INI_B$ which follows as $II_BI_C \sim IBC$

Similar solution with @math_comb01 Let $AI$, $BI$, $CI$ intersect $EF$ at $H$, $K$, $L$, respectively. Let $\angle ABI = \angle CBI = \beta$ and $\angle ACI = \angle BCI = \theta$. Then $\angle AFE = \angle AEF = \beta + \theta$. Thus, $\angle ELC = \beta$. Since $\angle ELC = \angle ABI = \beta$, thus $I$, $L$, $F$, $B$ are concyclic. Hence, $\angle ILB = \angle IFB = 90^\circ$. Similarly, $\angle IKC = 90^\circ$. This implies $B$, $L$, $K$, $C$ lie on a circle with center $M$, so $ML = MK$. $MG \perp EF$ and $ML = MK$ imply $GL = GK$. Since $\angle KLI = \angle CBI$, triangles $\triangle KLI \sim \triangle CBI$. In these triangles, $IH$ and $ID$ are altitudes, and $IG$ and $IM$ are medians, so $\angle DIM = \angle HIG = \angle IGM$. This implies that line $ID$ is tangent to the circumcircle $(MGI)$.

Solution with some calculations: Let $AI$ intersect $EF$ and $BC$ at points $H$ and $N$, respectively. Since $AH \perp EF$, it follows that $GM \parallel AN$. The problem asks us to show that $\angle MGI = \angle MID$. This implies $\angle HGI = \angle DMI$ and $\triangle HGI \sim \triangle DMI$. Let $\angle BAN = \alpha$, $\angle ACB = \theta$, $BC = a$, $AC = b$, $AB = c$, $u = \frac{a+b+c}{2}$, and $ID = r$. For ease of drawing, assume $b > c$. Since $\angle IFH = \angle FAI = \alpha$, we have $IH = IF \sin \alpha = r \sin \alpha$. Thus, the similarity ratio of triangles $HGI$ and $DMI$ is $\sin \alpha$. The question turns into proving $$ \dfrac{HG}{DM} = \sin \alpha \qquad (1)$$ Let $J$ be the foot of the perpendicular from $N$ to $MG$. Then $HG = NJ$ and $\sin \angle JMN = \sin (\alpha+\theta) = \frac{NJ}{NM} = \frac{HG}{NM}$. With $DM = \frac{a}{2} - (u - b) = \frac{b-c}{2}$ and $NM = BM - BN = \frac{a}{2} - \frac{ac}{b+c} = \frac{a(b-c)}{2(b+c)}$, we obtain $$ \frac{HG}{DM} = \frac{NM \sin (\alpha + \theta)}{DM} = \frac{a}{b+c} \cdot \sin (\alpha + \theta) \qquad (2)$$. By expressing sine ratios in triangle $ABN$, $\frac{\sin \angle BAN}{\sin \angle ANB} = \frac{\sin \alpha}{\sin (\alpha+\theta)} = \frac{BN}{AB} = \frac{NC}{AC} = \frac{BN+NC}{AB+AC} = \frac{a}{b+c}$, we substitute this result into $(2)$ to derive $(1)$.

Same idea with my previous solution but with less calculation. Let line $AI$ intersect $EF$ at $H$, $BC$ at $N$, and the circumcircle of triangle $ABC$ at $K$. Since $\angle CBK = \angle CAN = \angle BAN$, we have $\triangle KBN \sim \triangle KAB$. $$\frac{KN}{KB} = \frac{BN}{AB} = \frac{IN}{AI}$$ Since $\angle IBK = \angle IBN + \angle NBK = \angle IBA + \angle BAI = \angle BIK$, we find that $KI = KB$. $KM \perp BC$ and $IN \perp BC$ imply $$\frac{NM}{DM} = \frac{KN}{KI} = \frac{KN}{KB} = \frac{IN}{AI} \qquad (1)$$ Let $J$ be the foot of the perpendicular from $N$ to $MG$. Then $NJ = HG$. Since $\triangle INL \sim \triangle NJM$, we have $$\frac{HG}{NM} = \frac{NJ}{NM} = \frac{ID}{IN} \qquad (2)$$ By Euclid, $HI \cdot AI = IF^2 = ID^2$. Multiplying equations $(1)$ and $(2)$ side by side, we obtain $$\frac{HG}{DM} = \frac{ID}{AI} = \frac{HI}{ID}$$This implies that $\triangle HIG \sim \triangle DIM$. Therefore, $\angle DIM = \angle HIG = \angle IGM$, which means $ID$ is tangent to the circle $(MGI)$.

By Iran lemma, if $BI \cap EF = U$, then $\angle BUC = 90^{\circ}$. Analogously for $CI \cap EF = V$ we have $\angle BVC = 90^{\circ}$. Hence $BVUC$ is cyclic with diameter $BC$, in particular $MU = MV$ and from $MG \perp EF \equiv UV$ we obtain that $G$ is the midpoint of $UV$. Hence $\triangle BIC \sim \triangle VIU$ and the corresponding medians $IM$ and $IG$ yield $\angle IMB = \angle IGV$. Now from $\angle IDM = \angle MGV = 90^{\circ}$ we get $\angle DIM = \angle IGM$ and hence the desired tangency follows.

Set $(DEF)$ as the unit circle then::: $|d|_{cyc}=1$ Since $B$ and $C$ are the $\cap $ points of tangencies we get::: $b=\frac{2df}{d+f}$ $b=\frac{2ed}{e+d}$ Then $m=\frac{df}{d+f}+\frac{ed}{e+d}$ And $j=0$ where $j=Cor(I)$ Since both $E$ and $F$ lie on the unit circle we get::: $g=\frac{1}{2}(\frac{df}{d+f}+ \frac{ed}{d+e}+e+f-\frac{ef}{d+f}-\frac{ef}{d+e}$ We just want to prove that $O_{MIG}I \bot ID$ where $O_{MIG}$ is the center of $(MIG)$ What makes this question a worthy G5 is that when evaluating $O$ thing get very but very messy…