Let $P=\overline{AB}\cap\overline{CD}$, and let $Q=\overline{AD}\cap\overline{BC}$. WLOG, assume that $Q,B,C$ and $P,D,C$ lie in these orders. Then since $E$ and $G$ are the incenter and $Q$-excenter of $\triangle QAB$ and $\triangle QBD$ respectively, it follows that $Q,E,G$ are collinear. Similarly, $P,F,H$ are collinear. Let these two lines intersect at $I$. We will now show that $I$ and $T$ are isogonal conjugates with respect to quadrilateral $EFGH$. Notice that $\overline{ET}$ and $\overline{EI}$ are isogonal in $\angle E$, as $\overline{EI}$ passes through the circumcenter of $\triangle EAB$ by the incenter lemma. Similarly, $\{\overline{FT},\overline{FI}\}$, and $\{\overline{GT},\overline{GI}\}$ are also isogonal in $\angle F$ and $\angle G$. If $d(X,l)$ denotes the distance of point $X$ from line $l$, we have \[\frac{d(I,EH)}{d(I,EF)}=\frac{d(T,EF)}{d(T,EH)}\]and similarly for $F$ and $G$. Multiplying them gives us \[\frac{d(I,EH)}{d(I,GH)}=\frac{d(T,GH)}{d(T,EH)}\]which shows that $\overline{HI}$ and $\overline{HT}$ are isogonal in $\angle H$ as well. Therefore, $I$ and $T$ are isogonal conjugates. But this implies that \[\angle EIF+\angle GIH=180^\circ \Longrightarrow EG\perp FH.\]The rest can be done with angle chasing, but there is an easier way to see it. Notice that $\overline{BC}$ is obtained by reflecting $\overline{AD}$ over $\overline{QI}$. Since $\overline{PI}$ is perpendicular to $\overline{QI}$, reflecting $\overline{AD}$ over $\overline{PI}$ gives a line with the same direction as $\overline{BC}$. i.e. $AD$ and $BC$ are anti-parallel in $\angle P$, and the conclusion follows. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(150); pair B = dir(200); pair C = dir(340); pair D = dir(70); pair Q = extension(A,D,B,C); pair P = extension(A,B,C,D); pair E = incenter(Q,A,B); pair H = incenter(P,A,D); pair G = 2*circumcenter(incenter(Q,D,C),D,C)-incenter(Q,D,C); pair F = 2*circumcenter(incenter(P,B,C),B,C)-incenter(P,B,C); pair I = extension(E,G,F,H); pair T = extension(E,foot(E,A,B),F,foot(F,B,C)); draw(A--B--C--D--cycle, black+1); draw(A--Q); draw(B--Q); draw(A--P); draw(D--P); draw(Q--G); draw(P--F); draw(circumcircle(A,B,C), mydash); draw(E--F--G--H--cycle, royalblue+1); draw(E--T, mydash); draw(F--T, mydash); draw(G--T, mydash); draw(H--T, mydash); dot("$A$", A, dir(A)); dot("$B$", B, dir(225)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(135)); dot("$F$", F, dir(270)); dot("$G$", G, dir(0)); dot("$H$", H, dir(30)); dot("$I$", I, dir(135)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(180)); dot("$T$", T, dir(60)); [/asy][/asy]

Claim: Let the perpendicular from $F$ to $BC$ and $E$ ta $AB$ meet at $I$.Then unconditionally $EFGH$ is cyclic quadrilateral and $IE=IF$. Proof. $\angle FBC=90^{\circ}-\frac{\angle B}{2}$.Similarly $\angle EBK=90^{\circ}-\frac{\angle B}{2}$.Since $IF\perp BC$ and $IE\perp AB $ hence $\angle IEK=\angle IFE\implies IE=IF$. Concyclicity of $EFGH$ can be seen by observing that $\angle BFC= \frac{\angle B}{2}+\frac{\angle C}{2}$ and $\angle AHD= \frac{\angle A}{2}+\frac{\angle D}{2}$. So $\angle E+\angle H=1/2(\angle B+\angle C+\angle A+\angle D)=180^{\circ}$ $\blacksquare$ Now it is given that perpendicular from $G$ to $CD$ passes through $I$ which means $IE=IF=IG$.Hence,$I$ is the center of $\odot(EFGH)$.So $IH=IE$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.cm); 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 = 4., xmax = 17., ymin = -9., ymax = 5.; /* image dimensions */ draw((10.14376937909799,-3.6415989827081208)--(10.146249242070045,-4.054496167555314)--(10.559146426917238,-4.052016304583258)--(10.556666563945182,-3.6391191197360655)--cycle, linewidth(0.4)); draw((6.89135964938347,-1.4889448438771833)--(6.848514014259544,-1.8996204902526712)--(7.259189660635033,-1.9424661253765962)--(7.302035295758958,-1.5317904790011085)--cycle, linewidth(0.4)); draw(arc((7.08,-3.66),0.5839353302808407,84.04390064617643,132.19400765222179)--(7.08,-3.66)--cycle, linewidth(0.4)); draw(arc((7.08,-3.66),0.5839353302808407,-47.8059923477782,0.34411465826710574)--(7.08,-3.66)--cycle, linewidth(0.4)); draw(arc((13.74,-3.62),0.5839353302808407,140.98887432860377,180.3441146582671)--(13.74,-3.62)--cycle, linewidth(0.4)); draw(arc((10.579979669180178,-7.520751141362556),0.5839353302808407,92.83876732255845,132.1940076522218)--(10.579979669180178,-7.520751141362556)--cycle, linewidth(0.4)); /* draw figures */ draw((7.603780962826103,1.3604433857877372)--(7.08,-3.66), linewidth(0.8)); draw((7.08,-3.66)--(13.74,-3.62), linewidth(0.8)); draw((13.74,-3.62)--(12.62,1.82), linewidth(0.8)); draw((12.62,1.82)--(7.603780962826103,1.3604433857877372), linewidth(0.8)); draw((10.021221141248793,3.7476274186332916)--(4.9259298874607556,-1.2838922989204533), linewidth(0.8)); draw((10.579979669180178,-7.520751141362556)--(4.9259298874607556,-1.2838922989204533), linewidth(0.8)); draw((10.021221141248793,3.7476274186332916)--(16.072437338631424,-0.7408230775962321), linewidth(0.8)); draw((16.072437338631424,-0.7408230775962321)--(10.579979669180178,-7.520751141362556), linewidth(0.8)); draw(circle((10.541194481299426,-1.8746316629721598), 5.646252691392618), linewidth(0.8) + dotted); draw((4.9259298874607556,-1.2838922989204533)--(10.546042644552367,-1.8702365408322306), linewidth(0.8)); draw((10.546042644552367,-1.8702365408322306)--(10.579979669180178,-7.520751141362556), linewidth(0.8)); draw(circle((10.396958032139006,-1.468512351144544), 3.9755287323898796), linewidth(0.4) + dotted + blue); draw((10.546042644552367,-1.8702365408322306)--(16.072437338631424,-0.7408230775962321), linewidth(0.8)); draw((10.021221141248793,3.7476274186332916)--(10.546042644552367,-1.8702365408322306), linewidth(0.8)); draw((10.021221141248793,3.7476274186332916)--(10.579979669180178,-7.520751141362556), linewidth(0.8)); draw((10.246208988413184,-0.7896603479314981)--(7.08,-3.66), linewidth(0.8)); draw((10.246208988413184,-0.7896603479314981)--(13.74,-3.62), linewidth(0.8)); draw(circle((10.41309432879668,-4.155205744647027), 3.3696804795392694), linewidth(0.8)); draw(arc((13.74,-3.62),0.5839353302808407,140.98887432860377,180.3441146582671), linewidth(0.4)); draw(arc((13.74,-3.62),0.49634503073871455,140.98887432860377,180.3441146582671), linewidth(0.4)); draw(arc((13.74,-3.62),0.40875473119658845,140.98887432860377,180.3441146582671), linewidth(0.4)); draw(arc((10.579979669180178,-7.520751141362556),0.5839353302808407,92.83876732255845,132.1940076522218), linewidth(0.4)); draw(arc((10.579979669180178,-7.520751141362556),0.49634503073871455,92.83876732255845,132.1940076522218), linewidth(0.4)); draw(arc((10.579979669180178,-7.520751141362556),0.40875473119658845,92.83876732255845,132.1940076522218), linewidth(0.4)); /* dots and labels */ dot((7.603780962826103,1.3604433857877372),linewidth(1.pt) + dotstyle); label("$A$", (7.430215407467306,1.4673922376765536), NE * labelscalefactor); dot((7.08,-3.66),linewidth(1.pt) + dotstyle); label("$B$", (6.768422033149019,-4.041064377972716), NE * labelscalefactor); dot((13.74,-3.62),linewidth(1.pt) + dotstyle); label("$C$", (13.892433062575275,-4.002135355953993), NE * labelscalefactor); dot((12.62,1.82),linewidth(1.pt) + dotstyle); label("$D$", (12.705097891004232,1.856682457863781), NE * labelscalefactor); dot((4.9259298874607556,-1.2838922989204533),linewidth(1.pt) + dotstyle); label("$E$", (4.47160973404438,-1.4912134357463758), NE * labelscalefactor); dot((10.579979669180178,-7.520751141362556),linewidth(1.pt) + dotstyle); label("$F$", (10.66132423502129,-8.07021815691052), NE * labelscalefactor); dot((16.072437338631424,-0.7408230775962321),linewidth(1.pt) + dotstyle); label("$G$", (16.15031633966119,-0.6931684843625593), NE * labelscalefactor); dot((10.021221141248793,3.7476274186332916),linewidth(1.pt) + dotstyle); label("$H$", (9.980066349693644,3.8809916028373643), NE * labelscalefactor); dot((10.546042644552367,-1.8702365408322306),linewidth(1.pt) + dotstyle); label("$I$", (10.739182279058737,-2.2114003430927465), NE * labelscalefactor); dot((10.246208988413184,-0.7896603479314981),linewidth(1.pt) + dotstyle); label("$L$", (9.902208305656197,-0.6347749513344751), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: The internal angle bisector of $\angle B$ and $\angle C$ cut at $HF$. Proof. .Let the 2 internal bisector meet each other at $L$. We may assume that $BA$ and $CD$ are not parallel.(If they are parallel the claim can be check easily)Also we may assume $BA$ and $CD$ meet at $X$ so that $X$ and $BC$ lies in the opposite side of $AD$.The other case is analogus.2\angle So observe that $H$ is the incenter of $XAD$ ,$L$ is the incenter of $XBC$ and $F$ is the $X$-excenter of $XBC$.So $H,L,F$ lies on the angle bisector of $\angle BXC$. $\blacksquare$ Now observe that $LBFC$ is concyclic.Indeed $\angle LBF=\angle LCF=90$.This gives us $\angle LFB=\frac{\angle C}{2}$. Now note that,$\angle HIE=2\angle HFE=2\angle LFE=\angle C$. So, $$180^{\circ}=\angle IEH+\angle IHE+\angle HIE=2\angle IEH+\angle HIE=2[90^{\circ}-\angle EAK]+\angle C=\angle A+\angle C$$Hence $ABCD$ is cyclic quadrilateral. $\blacksquare$

Let $AB\cap CD=Q,\ AD\cap BC=P$ and internal angle bisectors of $A,B$ meet at $I_1$. Now, as $EHF$ and $QAB$ are orthologic, we have that the perpendicular from $Q$ to $AB$ passes through $I_1$ but observe that this is the angle bisector of $\angle AQB$. Thus, $EI_1\perp HF$. But, in $\Delta EAB$, $EI_1$ passes through circumcenter of $AB$. Thus, if we consider reflection of $EABI_1$ across angle bisector of $AEB$ and scale, we get that we can get $HF$. Thus, $ABHF$ is cyclic. Similarly, $CDHF$ are cyclic. But observe that $AB,CD,HF$ concur at $P$ as $H,F$ lie on the angle bisector of $APD$. But then by power of point from $P$, we get that $ABCD$ is cyclic and we are done.