Rather easy compared to previous IGO Advanced P5s. The cases where two opposite sides of $ABCD$ are parallel are easily dealt with. Let $O = AB \cap CD$. Then $OE^2 = OA \cdot OB = OC \cdot OD = OF^2$, so $OE = OF$. Reflect $E$ through $O$ onto $E'$, and notice that $OE^2 = OC \cdot OD$ implies $(C, D; E, E') = -1$. Because $\angle EFE' = 90^{\circ}$ (which follows from $OE = OF = OE'$) it follows that $FE$ bisects $\angle CFD$ and analogously $EF$ bisects $\angle AEB$. It then follows easily that $G$ and $H$ are symmetric about $EF$. Now let $I_1, I_2, I_3$ and $I_4$ be the incenters of $AGF, DGE, CHE, BHF$ respectively. Then $I_1I_2$ and $I_3I_4$ are the external bisectors of angles $EGF$ and $EHF$ respectively, and by symmetry about $EF$ these lines intersect at a (possibly ideal) point $X \in EF$. Finally, we may angle chase to find that $E, I_1, I_2, F$ and $E, I_3, I_4, F$ are quadruples of concyclic points. If $I_1I_2$ is parallel to $I_3I_4$ then we may easily conclude by symmetry about the perpendicular bisector of $EF$. Otherwise by Power of a Point from $X$ we have $XI_1 \cdot XI_2 = XE \cdot XF = XI_3 \cdot XI_4$, so $I_1, I_2, I_3, I_4$ are concyclic, as desired.

Let $P = \overline{AB}\cap\overline{CD}$, $\omega$ be the circle with centre $P$ and radius $PE$. Let $I_1, I_2, I_3, I_4$ be the incentre of triangles $AGF,BHF,CHE,DGE$. $\overline{I_1I_4}$ and $\overline{I_2I_3}$ are the external angle bisectors of $\angle EGF$ and $\angle EHF$, respectively. Note that $PE = PF$ since, \[PE^2 = PA\cdot PB = PC\cdot PD = PF^2.\] Also, due to above, $(A, B)$ and $(C, D)$ are inverses with respect to $\omega$. So, $\omega$ is an Appollonius circle of $(A, B)$ and $(C, D)$. Therefore, $\overline{EF}$ bisects $\angle AEB$ and $\angle CFD$. This implies $(\overline{FD}, \overline{FC})$ and $(\overline{EA}, \overline{EB})$ are reflections over $\overline{EF}$. Thus, $(G, H) \equiv (\overline{FD}\cap\overline{EA},\overline{FC}\cap\overline{EB})$ are reflections over $\overline{EF}$. Therefore, $(\overline{I_1I_4}, \overline{I_2I_3})$ are reflections over $\overline{EF}$. We conclude that $\overline{EF}, \overline{I_1I_4}, \overline{I_2I_3}$ are concurrent (or all parallel but we ignore this case for the sake of simplicity). Now note that $I_1I_4FE$ is cyclic. Since, \[2\angle I_4I_1F = 180^{\circ} + \angle GAF = 360^{\circ} - \measuredangle DEB = 2(180^{\circ} - \angle I_4EF).\]Similarly, $I_2I_3EF$ is cyclic. Let $K = \overline{EF}\cap\overline{I_1I_4}\cap\overline{I_2I_3}$, \[KI_1\cdot KI_4 = KE\cdot KF = KI_2\cdot KI_3.\]Thus, $I_1I_2I_3I_4$ is cyclic.

The following observation makes the angles in the problem tractable. Claim: We have that $EF$ is the angle bisector of $\angle AEB$ and similarly $FE$ is the angle bisector of $\angle DFC$. As a corollary, the quadrilateral $FGEH$ is a kite. Proof: Let $P=AB\cap CD$. Note that \[PE^2=PF^2=PA\cdot PB=PC\cdot PD.\]Let $E'$ be the reflection of $E$ over $P$, so $PE=PF=PE'$. Therefore, $P$ is the circumcenter of $\triangle EFE'$, so $FE'\perp FE$. However, we also have $(EE';CD)=-1$, so we have that $FE$ and $FE'$ are the internal and external angle bisectors of $\angle DFC$, as desired. $\blacksquare$ Let $W$, $X$, $Y$, $Z$ denote the incenters of triangles $AGF,BHF,CHE,DGE$ respectively. Note that $XY$ is the external angle bisector of $\angle FHE$ and $WZ$ is the external angle bisector of $\angle FGE$, so they concur on $EF$ by symmetry since $FHEG$ is a kite. Let the concurrency point be $T$. Note that $\angle AFG=\angle ECH$ and $\angle GAF=\angle HEC$ due to the tangencies, so $\triangle GAF\sim\triangle EHC$. Similarly, we have $\triangle GDE\sim\triangle HFB$. Label their angles by $\alpha$, $\beta$, $\gamma$, $\delta$ as shown in the diagram. Note that by the ratio lemma, we have \[\frac{TY}{TH}=\frac{EY}{EH}\cdot\frac{\sin\angle TEY}{\sin\angle TEH}=\frac{EY}{EH}\frac{\cos\beta}{\sin\angle TEH},\]so by a similar argument, we get \[\frac{TX\cdot TY}{TH^2}=\frac{EY\cdot FX}{EH\cdot FH}\frac{\cos\beta\cos\gamma}{\sin\angle FEH\sin\angle EFH}\]and \[\frac{TW\cdot TZ}{TG^2}=\frac{FW\cdot EZ}{EG\cdot FG}\frac{\cos\alpha\cos\delta}{\sin\angle FEG\sin\angle EFG}.\]Letting $r_W$, $r_X$, $r_Y$, $r_Z$ denote the inradii of the respective triangles, we see that \begin{align*} \frac{TX\cdot TY}{TW\cdot TZ} &=\frac{EY\cdot FX}{FW\cdot EZ}\frac{\cos\beta\cos\gamma}{\cos\alpha\cos\delta} \\ &= \frac{\frac{r_Y}{\sin\alpha}\cdot\frac{r_X}{\sin\delta}}{\frac{r_W}{\sin\gamma}\frac{r_Z}{\sin\delta}}\frac{\cos\beta\cos\gamma}{\cos\alpha\cos\delta} \\ &= \frac{r_Xr_Y}{r_Wr_Z}\frac{\sin(2\beta)\sin(2\gamma)}{\sin(2\alpha)\sin(2\delta)}. \end{align*}Note that the ratio of similarity in $\triangle GAF\sim\triangle EHC$ is the ratio of their circumradii, or \[\frac{\frac{FG}{\sin(2\alpha)}}{\frac{EH}{\sin(2\gamma)}}.\]Thus, \[\frac{r_Y}{r_W}=\frac{\sin(2\alpha)}{\sin(2\gamma)}\cdot\frac{EH}{FG}\]and similarly \[\frac{r_X}{r_Z}=\frac{\sin(2\delta)}{\sin(2\beta)}\cdot\frac{FH}{EG},\]so multiplying them shows that \[TX\cdot TY=TW\cdot TZ,\]proving that $W,X,Y,Z$ are concyclic. Remark: An alternative much faster way to finish is to note that $FXYE$ and $FWZE$ are cyclic by some angle chasing with $\alpha$, $\beta$, $\gamma$, $\delta$, so we have \[TW\cdot TZ=TE\cdot TF=TX\cdot TY,\]as desired.

Here's yet another finish using pure angles: Let the four incenters in the problem be $I_A, I_B, I_C, I_D$, and let $J = \overline{AI_A}\cap \overline{BI_B}$ and $K = \overline{CI_C}\cap \overline{DI_D}$ be the incenters of $\triangle ABE$ and $\triangle CDF$. [asy][asy] size(200); defaultpen(fontsize(10pt)); pair A, B, C, D, E, F, G, H, T, O, X, IA, IB, IC, ID, J, K; A = dir(140); B = dir(70); C = dir(330); D = dir(210); T = extension(A, B, C, D); O = (0,0); X = IP(CP(midpoint(T--O), T), circumcircle(A, B, C), 0); E = IP(CP(T, X), C--D); F = IP(CP(T, X), A--B); G = extension(A, E, D, F); H = extension(B, E, C, F); IA = incenter(A, F, G); IB = incenter(B, F, H); IC = incenter(C, H, E); ID = incenter(D, E, G); J = extension(A, IA, B, IB); K = extension(C, IC, D, ID); draw(A--B--C--D--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--B^^C--F--D, lightblue); draw(IA--IB--IC--ID--cycle, purple); draw(circumcircle(IA, IB, IC), dotted+magenta); draw(G--H^^E--F, heavygreen+dashed); draw(C--K--D, heavycyan); draw(A--J--B, heavycyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(270)); dot("$F$", F, dir(100)); dot("$G$", G, dir(180)); dot("$H$", H, dir(0)); dot("$I_A$", IA, dir(100)); dot("$I_B$", IB, dir(110)); dot("$I_C$", IC, dir(270)); dot("$I_D$", ID, dir(270)); dot("$J$", J, dir(220)); dot("$K$", K, dir(60)); [/asy][/asy] Claim: Line $EF$ bisects both $\angle AEB$ and $\angle CFD$. Proof. Let $T = \overline{AB}\cap \overline{CD}$ and let $E'$ be the reflection of $E$ over $T$. Since $$TE^2 = TA\cdot TB = TC\cdot TD = TF^2,$$we see that $\odot(T, TE)$ is orthogonal to $(ABCD)$ and thus $(E'E; CD) = -1$. Now $\overline{FE}$ bisects $\angle CFD$ since $\angle EFE' = 90^\circ$; by the same argument $\overline{EF}$ bisects $\angle AEB$. $\blacksquare$ By the claim, $J$ and $K$ lie on $\overline{EF}$. Claim: We have $\triangle KI_CE\sim \triangle FI_AJ$ and $\triangle KI_DE\sim \triangle FI_BJ$. In particular, we have $KI_CEI_D\sim FI_AJI_B$. Proof. Observe that \begin{align*} \angle KI_CE &= \angle ECH/2+\angle HEC/2 = \angle GFA/2 + \angle FAG/2 = \angle FI_AJ \\ \angle KEI_C &= \angle AEB/2+\angle HEC/2 = \angle AEB/2 + \angle EAB/2 = \angle FJI_A, \end{align*}which proves the first similarity. The other similarity follows analogously. $\blacksquare$ The similar quadrilaterals imply $$\angle(EF, I_AI_B) = \angle(EF, I_CI_D)\implies \angle(I_AI_B, AB) = \angle(I_CI_D, CD)$$since $\angle EFA = \angle FED$. Furthermore, since $\triangle AFG\sim \triangle ECH$, we have $$\angle(I_AI_D, AB) = \angle(I_BI_C, CD).$$Together with $ABCD$ cyclic, these two angle conditions imply $I_AI_BI_CI_D$ cyclic, as desired.

a bit easy for p5 but still nice let $R=AB \cap CD$ and the incenters be $I_a,I_b,I_c,I_d$ claim:$FE$ bisects $\angle CFD$ proof: note that $RE^2=RC.RD=RA.RB=RF^2$ $\angle EFD +\angle FDE =\angle FER=\angle EFR = \angle RFE+\angle CFE$ $\blacksquare$ now a simple angle chasing gives $FI_bI_cE$ and $FI_aI_dE$ are both cyclic note that $I_bI_C , I_aI_d , EF$ are concurrent in the $F-$ excenter in $\triangle FGH$ let it $T$ $TI_c.TI_b=TE.TF=TI_a.TI_d$ and we win

Let $I_1,I_2,I_3,I_4$ denote the incenters of $\triangle AGF,\triangle BHF,\triangle CHE,\triangle DGE$, respectively. Let $M=AI_1\cap BI_2$ and $N=CI_3\cap DI_4$. Claim 01. $M,N$ lie on $EF$. Proof. Note that $M,N$ are the incenters of $\triangle DFC$ and $\triangle ABE$, respectively. Hence, we would like to show that $EF$ bisects $\angle AEB$, indeed, note that $CE$ is tangent to $(AEB)$, hence $\measuredangle CEB=\measuredangle EAB$ and also $\measuredangle CEF=\measuredangle EFB$, therefore $\measuredangle BEF=\measuredangle FEA$. Similarly, $FE$ bisects $\angle DFC$. Claim 02. $I_1MNI_4$ and $MNI_2I_3$ are cyclic. Proof. This is just angle chase, \begin{align*} \measuredangle I_1MN=\measuredangle AME=90^\circ+\measuredangle ABE=90^\circ+\measuredangle GED=\measuredangle GI_4D=\measuredangle NI_4G, \end{align*}where the last inequality holds as $I_1,G,I_4$ all lie on the interior angle bisector of $\angle AGF$. Similarly, $MNI_2I_3$ is cyclic. Claim 03. $I_1I_4$, $I_2I_3$ and $MN$ are concurrent. Proof. Note that $\measuredangle EGF=\measuredangle EDF+\measuredangle AED=\measuredangle FBE+\measuredangle HFB=\measuredangle FHE$. Therefore if $P=GI_1\cap MN$, then $PH$ is the reflection of $PG$ over $MN$, hence $HI_2$ passes through $P$. Thus, we have $PI_1\cdot PI_4=PM\cdot PN=PI_2\cdot PI_3$. We are done. [asy][asy] import olympiad; size(10cm);defaultpen(fontsize(10pt)); pair O,A,B,C,D,X,E,F,I1,I2,I3,I4,G,H,M,N,P; O=(0,0);B=dir(78);A=dir(120);C=dir(30);D=dir(225);X=extension(A,B,C,D);E=intersectionpoints(circle(X,abs(sqrt(abs(C-X)*abs(D-X)))),C--D)[0]; F=intersectionpoints(circle(X,abs(sqrt(abs(C-X)*abs(D-X)))),A--B)[0];G=extension(A,E,F,D);H=extension(F,C,B,E); I1=incenter(A,G,F);I2=incenter(H,B,F);I3=incenter(H,E,C);I4=incenter(E,G,D);P=extension(I1,I4,I2,I3);M=extension(A,I1,B,I2);N=extension(C,I3,D,I4); draw(A--B--C--D--cycle,red);draw(circumcircle(A,B,C),royalblue);draw(P--E,heavygreen);draw(P--I4,heavygreen);draw(P--I3,heavygreen); draw(D--N--C,orange);draw(A--M--B,orange);draw(circumcircle(I1,I2,I3),heavyblue+dashed);draw(A--E--B,grey+0.3);draw(D--F--C,grey+0.3); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); dot("$I_1$",I1,dir(I1)); dot("$I_2$",I2,dir(I2)); dot("$I_3$",I3,dir(I3)); dot("$I_4$",I4,dir(I4)); dot("$P$",P,dir(P)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); [/asy][/asy]

Similar angle chasing solution. Let $K,L,M,N$ be the incenters of $GAF,HBF,HCE,GDE$ respectively. Let $AB\cap CD=P$. Note that $PE^2=PA.PB=PC.PD=PF^2$ gives $PE=PF$. Claim: $FGE\cong FHE$. Proof: \[\measuredangle EGF=\measuredangle APD+\measuredangle EAP+\measuredangle PDF=\measuredangle BPD+\measuredangle PEB+\measuredangle CFP=\measuredangle FHE\]\[\measuredangle EFH=\measuredangle EFP-\measuredangle HFP=\measuredangle PEF-\measuredangle PDF=\measuredangle DFE=\measuredangle GFE\]Thus, we get the result.$\square$ Claim: $E,N,K,F$ are concyclic. Proof: \[\measuredangle KFE=\measuredangle KFG+\measuredangle GFE=\frac{\measuredangle FCD}{2}+\frac{GFH}{2}=\frac{\measuredangle FCD+180-\measuredangle AFG-\measuredangle HFB}{2}=90-\frac{\measuredangle HFB}{2}=180-\measuredangle ENK\]Hence $E,N,K,F$ are concyclic.$\square$ Similarily, we conclude that $E,M,L,F$ are concyclic. If $KN\cap LM=S,$ then $SG=SH$ and $EF$ is the perpendicular bisector of $GH$ so $S,E,F$ are collinear. $SN.SK=SE.SF=SM.SL$ thus, $K,L,M,N$ are concyclic as desired.$\blacksquare$