Use complex bash using a unit circle the incircle to prove that $G$ is the center of spiral sim sending $A_bB_c$ to $DE$ and all other cyclic/symetric variations, $G$ could be annoying as it is a little messy to deal with so use line equation instead of the average colinearity condition to compute it easier and apart from that it should not take more than 2 pages, now by some easy synthetic geo it gives that $\triangle C_aA_bB_c \cup G \sim \triangle FDE \cup G \sim \triangle C_bA_cB_a \cup G$ which lead some useful lenght propeties used for the next part.

By simple cross ratio projection on $I$ one can prove $D$ is midpoint of $A_bA_c$, now you want to prove that $A_bC_bA_cB_c$ is cyclic and cyclic variations of this, by proving that $ID$ is symedian in $\triangle C_bIB_c$, here the process is trig bashing with LoS spam using the ratios given in Step 1 until things eventually cancel out, this should not take more than 6 to 7 lines, if it does you are probably overcomplicating this.

Prove that the circles $(A_bB_cC_a), (A_cB_aC_b)$ are symetric w.r.t. $I$ which by symetry and Step 2 gives that it lies on the radax of the circles. This is done by considering $IB_c \cap (A_bB_cC_a)$ and then using reims and midbase to get that such point is just reflecting $A_c$ over $I$, after that repeat the same process for all points to prove this claim.)

Use linearity of Power of the point on $G$ to prove that it lies on the radax of the two circles in step 3 with a special technique involving bary coordinates, basically since $G=(p-a ; p-b ; p-c)$ in bary, you can use this by defining $f(X)$ as the difference of power of point in both circles of point $X$ and noticing that all you need is to prove that $(p-a)f(A)+(p-b)f(B)+(p-c)f(C)=0$ which can be done by some easy lenght chase (Basically first of all get rid of the $A_b, A_c$'s by involving $I$ into this and stuff eventually cancels out, this should not take more than 1 page.

Step 1 complex bash can be skipped by consider the midpoint of $EF$ and then finish angles using the fact that the 90 degrees in each perpendicular from like $D$ to $AG$ gives isogonalities, which can give you angle equalities that derive a one liner sine bash to prove similar isosceles triangles and then just spiral sim your way to the top. Final Step Lin PoP can be skipped by considering rotations over $G$ as there is chain of similar triangles, along with step 3 you also get that the circles are congruent and thus G is symetric on them by checking the lenghts after rotiations, giving a full synthetic sol for this, yay.

Let $\Gamma_1, \Gamma_2$ denote the circles $(A_bB_cC_a)$ and $(B_aC_bA_c)$ respectively. Denote by $M_a, M_b, M_c$ the midpoints of $EF, FD, DE$ respectively. Claim: $D$ is the midpoint of $A_bA_c$. Proof: We have \[\frac{DA_b}{DA_c} = \frac{DM_b \cos \angle A_cDM_c}{DM_c \cos \angle A_bDM_b} = \frac{DF \sin \angle EDG}{DE \sin \angle FDG} = 1 \]since $DG$ is the symmedian in $\triangle DEF {}$. Claim: $\angle B_aGB_c = \angle C_bGC_a$. Proof: Since triangles $GB_aB_c$ and $GC_bC_a$ are isosceles, it suffices to show that \[\frac{GE}{GF} = \frac{EB_a}{FC_a}\]To see the above, we note that \[\frac{GE}{GF} = \frac{\sin GFE}{\sin GEF} = \frac{\sin FC_aM_a}{\sin EB_aM_a} = \frac{EB_a}{FC_a}\]because $M_a$ is midpoint of $EF$. From the above 2 claims, we get that the $GA_bDA_c, GB_cEB_a, GC_aFC_b$ are similar figures. This implies that $G$ is the common spiral center between the triangles $A_bB_cC_a,DEF,A_cB_aC_b$ and since their circumcenters are $O_1,I,O_2$ we get that $GO_1IO_2$ is also similar to $GA_bDA_c$. Hence, $GO_1 = GO_2$ and $I$ is the midpoint of $O_1O_2$ giving that $IG$ is the perpendicular bisector of $O_1O_2$. $\square$

Denote $K = EF \cap BC$ then $(A_BA_C;DP_{\infty})=(BC;DK)=-1$ where we used the well known lemma that $KI \perp AD$

Introduce midpoints and then sin rule finishes

Spam Claim 2

length chase

it turns out that $I$ is actually the isodynamic point of both triangles