Lemma. $DE\perp OI_A$ (where $O$ is the circumcenter, $I_A$ is the $A$-excenter). Proof. Let $AOI_AP$ be a parallelogram, and let $P_B,P_C$ be the projections from $P$ onto $AB,AC$. It is easy to check that $\frac{AP_B}{AP_C} = \frac{AE}{AD}$ (using angle bisector theorem), so the lemma follows. Let $M_C,M_B$ be the midpoints of minor arcs $AB,AC$, and let the tangents at $M_B,M_C$ (to $(O)$) intersect at $T$. Then by Pascal on $BAACM_CM_B$, we get that $AA,M_BM_C,DE$ concur. Hence the pole of $DE$ is on $AT$, and by the lemma it is on $OI$. However, $T+(O)$ and $A+(I)$ are homothetic (call the center of homothety $X$), so $X=AT\cap OI$ is the pole of $DE$, done.

Lemma: $OI_A \perp DE$. Proof: Just use Carnot and length bash to show this result. $\blacksquare$ By Lemma, we have $OI_A$ is perpendicular bisector of $XY$, hence the tangents at $X,Y$ and $OI_A$ are concurrent, call that point $Z$ and let $N = OI_A \cap XY, X' = XO \cap (ABC)$. It then suffices to show $\frac{ZO}{ZI_A} = \frac{R}{r_a}$ where $r_a$ is the exradius and $R$ is the circumradius. Let $I_A,I_B,I_C$ be the excenters opposite to $A,B,C$ and let $S=I_AX \cap (ABC),T=I_AY\cap (ABC)$. Then PoP from $D$ and $E$ gives that $I_BI_CXIY$ lie on a circle. Also it is well known that in any triangle $ABC$, the homothety at $A$ with ratio $2$ takes the 9-point circle to the circle $(BHC)$ of $ABC$. ($H$ is the orthocenter of $ABC$.) Applying this to triangle $I_AI_BI_C$, we see that $S$ and $T$ are the midpoints of $I_AX$ and $I_AY$. Let $\alpha = \angle OXY , \beta = \angle OXT$. Then by angle chase we have $\angle OI_AX = \beta$ and $\angle OXI_A = 90^\circ - \alpha - 2\beta$. So we have $\frac{NO}{NI_A} = \frac{R\sin{\alpha}}{\frac{R\cos{\alpha}\cos{\beta}}{\sin{\beta}}}= \frac{\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}}$. We have $\frac{ZO}{ZI_A} \cdot \frac{NI_A}{NO} = (Z,N;O,I_A)\stackrel{X} = (X,Y;X',S) = \frac{2R}{2R\sin{(\alpha+\beta)}}\cdot \frac{2R\cos{\beta}}{2R\sin{\alpha}} = \frac{\cos{\beta}}{\sin{\alpha}\sin{(\alpha+\beta)}} \implies \frac{RO}{RI_A}=\frac{\sin{\beta}}{\sin{(\alpha+\beta)}\cos{\alpha}}$. By PoP , $I_AX \cdot I_AS = 2SX^2 = 8R^2\sin^2{(\alpha+\beta)} = OI_A^2 - R^2 = 2Rr_a \implies \frac{R}{r_a} = \frac{1}{4\sin^2{(\alpha+\beta)}}$. (We used Euler's formula.) Thus it suffices to show $\frac{\sin{\beta}}{\cos{\alpha}} = \frac{1}{4\sin{(\alpha+\beta)}}$. By LOS in $XOI_A$, $\frac{R}{4R \sin{(\alpha+\beta)}} = \frac{R}{I_AX} = \frac{\sin{\beta}}{\cos{\alpha}} \implies \frac{\sin{\beta}}{\cos{\alpha}}=\frac{1}{4\sin{(\alpha+\beta)}}$, thus we are done. $\square$

Let $\omega$ be the $A$-excircle. Let $P,Q,R$ be the inctouch points to $BC$, $CA$,$AB$ respectively and let $T$ be the touch point of the second tangent from $E$ to $\omega$. Redifine $F,G$ as the touch points of the common tangent. Lemma: $E$,$F$ lie on the polar of the centroid of $PQR$ w.r.t. $\omega$ . Proof: We wish to show that $R,M,T$ are collinear where $M$ is the midpoint of $PQ$. Let $RM$ meet $\omega$ at $T'$, by PoP $R, I_A,C,T'$ are concyclic so if $E'$ is the intersection of the tangent from $T'$, $\angle I_ACE' = 90$ meaning $E=E', T=T'$. Now move $A$ on the circumcircle, by poncelet's porism there exists $B'$, $C'$ such that $\omega$ is still the excircle of $AB'C'$. Let the centroid of $PQR$ be $K$, notice that $K$ is a fixed point as $A$ varies, since the circumcircle, the center of $NPC$ and itself are all fixed (to see that $NPC$ is fixed just invert from $I$). So we deduce that the locus of points $D,E$ are a line, the prove that this line is $FG$, pick $A$ to be $F$. Let the excircle and the circumcircle meet at $X,Y$, by the definition of poncelet's porism the tangent from $X$ to $\omega$ passes trough $G$ finishing the proof.