Just notice that letting $F$ be the intouch point on $BC$ and $F'$ be the reflection of $F$ through $I$ suffices that $A,F',E$ collinear; which can be done in many ways, homothety from incircle to excircle // complex bashing is also simple just define the unit circle as the incircle and the calculation is simple.

Solution. Let $X=\gamma\cap BC$, $Y=XI\cap\gamma$, $Z=A$-excircle $\cap BC$. We can note that $DZ=DX=DE\implies \angle XEZ=90^\circ =\angle XEY\implies A-Y-E-Z$ collinear. Now since $I$ is the midpoint of $XY$, and $D$ is the midpoint $XZ\implies ID\parallel IE$.$\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.137272727272734, xmax = 15.571818181818173, ymin = -3.5267768595041304, ymax = 12.145950413223142; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((-11.,10.)--(-10.,1.)--(3.,1.)--cycle, linewidth(0.8) + zzttqq); /* draw figures */ draw((-11.,10.)--(-10.,1.), linewidth(0.8) + zzttqq); draw((-10.,1.)--(3.,1.), linewidth(0.8) + zzttqq); draw((3.,1.)--(-11.,10.), linewidth(0.8) + zzttqq); draw(circle((-7.293965919477911,4.023357208508352), 3.0233572085083535), linewidth(0.8)); draw((-7.29396591947791,4.02335720850835)--(3.,1.), linewidth(0.8)); draw((-4.34689483613138,4.698235597510502)--(-3.5,1.), linewidth(0.8)); draw((-11.,10.)--(-4.34689483613138,4.698235597510502), linewidth(0.8) + red); draw((-7.29396591947791,4.02335720850835)--(-3.5,1.), linewidth(0.8) + red); draw((-4.34689483613138,4.698235597510502)--(0.2939658157318217,1.), linewidth(0.8) + red); draw((-7.293965919477913,7.046714417016706)--(-7.293965815731822,1.), linewidth(0.8) + zzttqq); draw((-4.34689483613138,4.698235597510502)--(-7.293965815731822,1.), linewidth(0.8)); /* dots and labels */ dot((-11.,10.),linewidth(4.pt) + dotstyle); label("$A$", (-10.905090909090921,10.177768595041323), NE * labelscalefactor); dot((-10.,1.),linewidth(4.pt) + dotstyle); label("$B$", (-10.06909090909092,0.4317685950413166), NE * labelscalefactor); dot((3.,1.),linewidth(4.pt) + dotstyle); label("$C$", (3.0869090909090775,1.1797685950413173), NE * labelscalefactor); dot((-7.29396591947791,4.02335720850835),linewidth(4.pt) + dotstyle); label("$I$", (-7.715090909090922,3.819768595041319), NE * labelscalefactor); dot((-3.5,1.),linewidth(4.pt) + dotstyle); label("$D$", (-3.645090909090922,0.47576859504131663), NE * labelscalefactor); dot((-4.34689483613138,4.698235597510502),linewidth(4.pt) + dotstyle); label("$E$", (-4.217090909090922,4.85376859504132), NE * labelscalefactor); dot((-7.293965815731822,1.),linewidth(4.pt) + dotstyle); label("$X$", (-7.473090909090922,0.49776859504131665), NE * labelscalefactor); dot((0.2939658157318217,1.),linewidth(4.pt) + dotstyle); label("$Z$", (0.2049090909090778,0.5197685950413167), NE * labelscalefactor); dot((-7.293965919477913,7.046714417016706),linewidth(4.pt) + dotstyle); label("$Y$", (-7.671090909090922,6.569768595041321), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Let $X$ be the foot of perpendicular from $I$ and $Y$ be the antipode of $X$ on incircle. Well known that $A-Y-E$ are collinear. Since $IE = IX$ and $XD = DE$, $IXDE$ is a kite hence the diagonals are perpendicular. Let $P$ be the intersection of the diagonals. $$\angle YEX = 90^\circ = IPX \implies IP \parallel YE \implies ID \parallel AE$$