Let $AD \cap BC = E, AB \cap CD = F$. Note $\angle BKL = \angle BAC = \angle BDC = \angle LMC$ so lines $AB$ and $CD$ are symmetric about $LN$, and hence $LN$ passes through $F$. Similarly $KM$ passes through $E$. Let $KB \cap ML = X, KL \cap MC = Y$ so by symmetry the self-intersecting quadrilateral $KXMY$ has an excircle $\Omega$. Then by Monge on $\omega_B, \omega_C, \Omega$ we have $KM$ passes through the exsimilicenter of $\omega_B, \omega_C$. Since this exsimilicenter also lies on $BC$, it is $E$. If $T$ is the insimilicenter of $\omega_B, \omega_C$ then $(I_BI_C;ET)=-1$ so by Ceva-Menelaus we have $FT$ passes through $I_AI_C \cap I_BI_D$. But by Monge $FT$ intersects $I_AI_C$ at the insimilicenter of $\omega_A, \omega_C$. So the insimilicenter of $\omega_A, \omega_C$ is $I_AI_C \cap I_BI_D$ and similarly this point is also the insimilicenter of $\omega_B, \omega_D$.

This problem was proposed by Burii.

Some ridiculous vector shenanigans. Still not sure why I chose to take this route. Let $AC=m$, $BD=n$, $AB=a$, $BC=b$, $CD=c$, $DA=d$. We use vectors, with capital letters representing the vectors. Since $KL=MN$ and $KL\parallel AC\parallel MN$, we must have $$\frac{BK}{KA}=\frac{BL}{LC}=\frac{DN}{NA}=\frac{DM}{MC}.$$ We also must have $$n\cdot \frac{AK}{AB}=m\cdot \frac{BK}{BA},$$so $\frac{AK}{BK}=\frac mn$. Hence, \[AK=\frac{m}{m+n}\cdot a, BK=\frac{n}{m+n}\cdot a, BL=\frac{n}{m+n}\cdot b, CL=\frac{m}{m+n}\cdot b,\]etc. Also, the rhombus' side length is $\frac{mn}{m+n}.$ Now, we have \[K=\frac{mB+nA}{m+n}, N=\frac{mD+nA}{m+n},\]etc. The incenter of $\triangle AKN$ is at \[I_A = \frac{A\cdot \frac{m}{m+n}\cdot n + K\cdot \frac{m}{m+n}\cdot d+N\cdot \frac{m}{m+n}\cdot a}{\frac{m}{m+n}\cdot (n+a+d)}=\frac{An+Kd+Na}{n+a+d}.\]Similarly, the incenter of $\triangle CML$ is at \[I_C=\frac{Cn+Mb+Lc}{n+b+c}.\] If $r_A=\operatorname{inradius}(\triangle AKN)$ and $r_C=\operatorname{inradius}(\triangle CML)$, we have that the common internal tangents of the two incircles meet at \[\frac{r_A\cdot I_C+r_C\cdot I_A}{r_A+r_C}.\]Since $\frac{r_A}{r_C}=\frac{[AKN]/\operatorname{perimeter}(AKN)}{[CML]/\operatorname{perimeter}(CML)},$ \[\frac{r_A}{r_C}=\frac{ad/(n+a+d)}{bc/(n+b+c)}=\frac{ad(n+b+c)}{bc(n+a+d)},\] so the intersection is at \begin{align*}&\frac{ad(n+b+c)\cdot \frac{Cn+Mb+Lc}{n+b+c}+bc(n+a+d)\frac{An+Kd+Na}{n+a+d}}{ad(n+b+c)+bc(n+a+d)}\\ =&\frac{adnC+bcnA+adbM+bcdK+adcL+bcaN}{(ad+bc)n+adb+adc+bca+bcd}\\ =&\frac{4[ABD]\cdot \text{circumradius}\cdot C+4[BCD]\cdot \text{circumradius}\cdot A+\text{symmetric}}{4[ABCD]\cdot \text{circumradius}+\text{symmetric}}\\ =&\frac{\overrightarrow{AC\cap BD}\cdot \text{symmetric constants}+\text{symmetric}}{\text{symmetric}} \end{align*}where $R$ is the circumradius and we used the fact that $\frac{abc}{4R}=[ABC]$ in triangles $ABC$. This expression is symmetric (if we replace $ABCD$ with $BCDA$), the other pair of tangents intersect at the same place, and we are done. $\blacksquare$

Diagram

Let $P=\overline{AD}\cap \overline{BC}$ and $Q=\overline{AD}\cap \overline{BC}$ and $R=\overline{AD}\cap \overline{BC}$. Let $I_A$ be the center of $\omega_A$, and similarly define $I_B,I_C,I_D$. Denote by $\mathbf{E}(\omega,\Omega)$ the exsimilcenter and $\mathbf{I}(\omega,\Omega)$ the insimilcenter of any two circles $\omega$ and $\Omega$. Claim 1: $K,M,P$ collinear, and furthermore, $\overline{PMK}$ bisects $\angle (\overline{AD},\overline{BC})$. Proof: We show $\overline{PK}$ bisects $\angle APB$ first. We will similar triangles created by the parallel lines: \[ \frac{BK}{BA} = \frac{KL}{AC} = \frac{KN}{BD}\cdot \frac{BD}{AC} = \frac{AK}{AB}\cdot \frac{BD}{AC} \implies \frac{AK}{BK} = \frac{AC}{BD}. \]Due to $ABCD$ being cyclic, we have $\triangle PCA\sim \triangle PDB$ and $\triangle PDC\sim \triangle PBA$. Hence \[ \frac{PA}{PB}=\frac{PC}{PD}=\frac{AC}{BD}. \]Putting it together, $PA/PB=KA/KB$, so by the Angle Bisector Theorem, $\overline{PK}$ bisects $\angle APB$. Similarly, we can show $\overline{PM}$ bisects $\angle DPC$. $\blacksquare$ Claim 2: We have $\mathbf{E}(\omega_A,\omega_D)=\mathbf{E}(\omega_B,\omega_C)=P$. Proof: This is the trickiest part of the proof. Let $L'=\overline{NM}\cap \overline{PB}$ and $N'=\overline{LM}\cap \overline{PA}$. Since $MNKL$ is a rhombus, $\overline{PMK}\perp \overline{LN}$, but since $\overline{PMK}$ bisects $\angle NPL$ by Claim 1, $\triangle PNL$ is isosceles. We have $\triangle PL'N\sim \triangle PCA$, so \[ \frac{PL'}{PN} =\frac{PC}{PA}=\frac{PD}{PB}=\frac{PN'}{PL},\]and since $PN=PL$ by tangency, we also have $PL'=PN'$. Hence $\overline{L'N'}\perp \overline{PMK}$. Hence $PL'MN'$ is a kite, so it has an incircle $\gamma$. Now, Monge's Theorem implies $\mathbf{E}(\omega_C,\omega_D)\in \overline{LN}$. But it also lies on $\overline{CD}$, the external tangent of $\omega_C$ and $\omega_D$. Hence $\mathbf{E}(\omega_C,\omega_D)=\overline{LN}\cap \overline{CD}=Q$, and the rest follow by symmetry. $\blacksquare$ A corollary of Claim 2 is that $Q=\overline{I_CID}\cap \overline{I_BI_A}$ and $P=\overline{I_AI_D}\cap \overline{I_BI_C}$. By Monge on $\omega_A, \omega_B, \omega_D$, we have $\mathbf{E}(\omega_A,\omega_D)=P$, $\mathbf{I}(\omega_B,\omega_D) := X$, and $\mathbf{I}(\omega_A, \omega_B):=Q_1$, the harmonic conjugate of $Q$ in $\overline{I_AI_B}$, are collinear. By Monge on $\omega_B,\omega_C,\omega_D$, we have $\mathbf{E}(\omega_C,\omega_D)=Q$, $\mathbf{I}(\omega_B,\omega_D) := X$, and $\mathbf{I}(\omega_B, \omega_C):=P_1$, the harmonic conjugate of $P$ in $\overline{I_BI_C}$, are collinear. By Ceva-Menelaus (or equivalently, Brokard) in $I_AI_BI_CI_D$, we have $\overline{QP_1} \cap \overline{PQ_1} = \overline{I_BI_D} \cap \overline{I_AI_C}$. Hence $X=\overline{I_BI_D} \cap \overline{I_AI_C}$, which is symmetric upon switching $B$ and $D$ with $A$ and $C$, so we are done.

Let \(F=\overline{AB}\cap\overline{CD}\) and \(G=\overline{AD}\cap\overline{BC}\). Claim: \(K\) and \(M\) lie on the bisector of \(\angle G\). Proof. Note that \(\frac{AK}{AB}=\frac{KN}{BD}\) and \(\frac{BK}{AB}=\frac{KL}{AC}\). Dividing, we have \[\frac{AK}{BK}=\frac{AC}{BD}=\frac{GA}{GB},\]and similarly for \(M\). \(\blacksquare\) Analogously \(F\), \(N\), \(L\) collinear. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=red; pen sec2=lightred; pen tri=heavygreen; pen qua=fuchsia+pink; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen sfil2=invisible; pen tfil=invisible; pair A,B,C,D,F,G,K,M,L,NN,IA,IB,IC,ID,IG; A=dir(150); B=dir(80); C=dir(-20); D=reflect( (0,0),(0,1))*C; F=extension(A,B,C,D); G=extension(A,D,B,C); K=extension(A,B,G,incenter(G,A,B)); L=extension(B,C,F,incenter(F,B,C)); M=extension(C,D,G,K); NN=extension(D,A,F,L); IA=incenter(A,NN,K); IB=incenter(B,K,L); IC=incenter(C,L,M); ID=incenter(D,M,NN); IG=extension(NN,IA,L,IB); draw( (-2K+3foot(IG,K,L))--K--(-2K+3foot(IG,K,NN)),tri); draw(F--(-.3F+1.3(reflect(IA,IB)*foot(IB,A,B))),sec2); filldraw(circle(IG,abs(IG-foot(IG,G,A))),sfil2,sec2); filldraw(K--L--M--NN--cycle,tfil,tri); filldraw(incircle(A,NN,K),sfil,sec); filldraw(incircle(B,K,L),sfil,sec); filldraw(incircle(C,L,M),sfil,sec); filldraw(incircle(D,M,NN),sfil,sec); draw(G--M,pri2); draw(F--L,pri2); filldraw(unitcircle,fil,pri); filldraw(A--B--C--D--cycle,fil,pri); draw(A--F--D,pri); draw(A--G--B,pri); dot("\(A\)",A,NW); dot("\(B\)",B,NE); dot("\(C\)",C,C); dot("\(D\)",D,SW); dot("\(F\)",F,W); dot("\(G\)",G,N); dot("\(K\)",K,N); dot("\(L\)",L,NE); dot("\(M\)",M,S); dot("\(N\)",NN,dir(150)); [/asy][/asy] Claim: \(F\) is the exsimilicenter of \(\omega_A\) and \(\omega_B\). Proof. Since \(\overline{GKM}\) is the perpendicular bisector of \(\overline{NL}\), by symmetry there is a circle \(\omega_G\) tangent to \(\overline{GN}\), \(\overline{GL}\), \(\overline{KN}\), \(\overline{KL}\). By Monge's theorem on \(\omega_A\), \(\omega_B\), \(\omega_G\), the exsimilicenter of \(\omega_A\), \(\omega_B\) lies on line \(LN\), which is sufficient. \(\blacksquare\) Analogously \(\omega_C\), \(\omega_D\) have exsimilicenter \(F\), and \(\omega_A\), \(\omega_D\) and \(\omega_B\), \(\omega_C\) have exsimilicenter \(G\). [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=lightblue; pen sec=lightred; pen tri=heavygreen; pen qua=fuchsia+pink+linewidth(1); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,D,F,G,K,M,L,NN,IA,IB,IC,ID,IG,X; A=dir(150); B=dir(80); C=dir(-20); D=reflect( (0,0),(0,1))*C; F=extension(A,B,C,D); G=extension(A,D,B,C); K=extension(A,B,G,incenter(G,A,B)); L=extension(B,C,F,incenter(F,B,C)); M=extension(C,D,G,K); NN=extension(D,A,F,L); IA=incenter(A,NN,K); IB=incenter(B,K,L); IC=incenter(C,L,M); ID=incenter(D,M,NN); IG=extension(NN,IA,L,IB); X=extension(IA,IC,IB,ID); draw(IA--IB--IC--ID--cycle,qua); draw(IA--F--ID,qua); draw(IA--G--IB,qua); draw(IA--IC,qua); draw(IB--ID,qua); filldraw(K--L--M--NN--cycle,tfil,tri); filldraw(incircle(A,NN,K),sfil,sec); filldraw(incircle(B,K,L),sfil,sec); filldraw(incircle(C,L,M),sfil,sec); filldraw(incircle(D,M,NN),sfil,sec); filldraw(unitcircle,fil,pri); filldraw(A--B--C--D--cycle,fil,pri); draw(A--F--D,pri); draw(A--G--B,pri); dot("\(F\)",F,W); dot("\(G\)",G,N); dot("\(I_A\)",IA,NW); dot("\(I_B\)",IB,E); dot("\(I_C\)",IC,SE); dot("\(I_D\)",ID,S); [/asy][/asy] This is enough to solve the problem. Let \(I_A\), \(I_B\), \(I_C\), \(I_D\) be the centers of \(\omega_A\), \(\omega_B\), \(\omega_C\), \(\omega_D\). To finish, observe: By Monge on \((\omega_A,\omega_D,\omega_C)\) and \((\omega_A,\omega_D,\omega_B)\), the insimilicenters of \((\omega_A,\omega_C)\) and \((\omega_B,\omega_D)\) lie on the line through \(F\) and the insimilicenter of \((\omega_A,\omega_D)\). By Monge on \((\omega_A,\omega_B,\omega_C)\) and \((\omega_A,\omega_B,\omega_D)\), the insimilicenters of \((\omega_A,\omega_C)\) and \((\omega_B,\omega_D)\) lie on the line through \(G\) and the insimilicenter of \((\omega_A,\omega_B)\). Hence the two insimilicenters coincide. Remark: [Generalization] The problem still holds if \(KLMN\) is any parallelogram.

The assumption that $KLMN$ is a rhombus can be weakened to the assumption that $KLMN$ is a parallelogram, as can be seen in the following solution. Let $I_1$, $I_2$, $I_3$, $I_4$ be the centers of $\omega_A$, $\omega_B$, $\omega_C$, $\omega_D$, respectively. Note that $$\angle I_1KN = \frac 12 \angle AKN = \frac 12 \angle ABD = \frac 12 \angle ACD = \frac 12 \angle NMD = \angle NMI_4.$$Similarly, $\angle I_2LK=\angle KNI_1$, $\angle I_3ML = \angle LKI_2$, $\angle I_4NM = \angle MLI_3$. Let $P$ and $Q$ be the points on $ML$ such that $\angle I_1KN = \angle I_3PL$ and $\angle I_2LK = \angle MQI_3$. Let $R$ be the point on $MN$ such that $\angle I_3ML = \angle NRI_4$. Let $S$ be the point on $KL$ such that $\angle I_4NM = \angle I_2SK$. The angle equalities clearly imply that $I_2KSL \sim I_3MLQ$. Hence $$\frac{QL}{LS} = \frac{QM}{LK} = \frac{QM}{MN}.$$Since $\angle QLS = \angle QMN$, it follows that $\triangle QLS \sim \triangle QMN$, and consequently points $Q$, $S$, $N$ are collinear. Similarly, $I_4RMN \sim I_3MPL$, so $$\frac{RM}{MP}=\frac{RN}{ML}=\frac{RN}{NK}.$$Since $\angle RMP = \angle RNK$, we obtain $\triangle RMP \sim \triangle RNK$ and therefore $R$, $P$, $K$ are collinear. Let $Z$ be the intersection of lines $QSN$ and $RPK$. Then, by definition of $P$ and $Q$, there exists a homothety with center $Z$ and negative scale which maps triangle $I_1KN$ into $I_3PQ$. This homothety maps the circle with center $I_1$ tangent to $KN$ to the circle with center $I_3$ tangent to $PQ$, i.e. it maps $\omega_A$ to $\omega_C$. This means that $Z$ is the insimilicenter of $\omega_A$ and $\omega_C$ and therefore their common internal tangents pass through $Z$. We prove analogously that the common internal tangents to $\omega_B$ and $\omega_D$ pass through $Z$. This finishes the proof.

Attachments:

[asy][asy] size(10cm); 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 = -3.16, xmax = 14.68, ymin = -4.94, ymax = 5.54; /* image dimensions */ /* draw figures */ draw((-0.26,0.06)--(2.8,-0.06), linewidth(0.8)); draw((4.008710587909909,2.753719729232903)--(2.8,-0.06), linewidth(0.8)); draw((0.948710587909908,2.873719729232904)--(-0.26,0.06), linewidth(0.8)); draw((0.948710587909908,2.873719729232904)--(4.008710587909909,2.753719729232903), linewidth(0.8)); draw((-1.29430082475954,1.790464841424615)--(0.78,-1.68), linewidth(0.8)); draw((0.78,-1.68)--(4.808930448090646,1.5511224385677451), linewidth(0.8)); draw((4.808930448090646,1.5511224385677451)--(3.204081378819364,3.962943537579509), linewidth(0.8)); draw((3.204081378819364,3.962943537579509)--(-1.29430082475954,1.790464841424615), linewidth(0.8)); draw(circle((1.7407897092296702,1.249403527881168), 3.0829404623709924), linewidth(0.8)); draw((3.204081378819364,3.962943537579509)--(0.78,-1.68), linewidth(0.8)); draw((-1.29430082475954,1.790464841424615)--(0.5864177373916837,1.2270249225329872), linewidth(0.8)); draw((3.403602911632138,1.7690666782594093)--(4.808930448090646,1.5511224385677451), linewidth(0.8)); draw((3.204081378819364,3.962943537579509)--(2.845903694445863,2.6528470238672526), linewidth(0.8)); draw((0.78,-1.68)--(1.1441169545779593,0.34324457692514415), linewidth(0.8)); draw((-1.29430082475954,1.790464841424615)--(4.808930448090646,1.5511224385677451), linewidth(0.8)); draw((-0.3565251960972488,1.509518912147643)--(4.108197259988662,1.6597951557426085), linewidth(0.8)); /* dots and labels */ dot((-0.26,0.06),dotstyle); label("$K$", (-1.02,0.12), NE * labelscalefactor); dot((2.8,-0.06),dotstyle); label("$N$", (2.76,-0.64), NE * labelscalefactor); dot((4.008710587909909,2.753719729232903),dotstyle); label("$M$", (4.22,2.86), NE * labelscalefactor); dot((0.948710587909908,2.873719729232904),linewidth(4pt) + dotstyle); label("$L$", (1.02,3.04), NE * labelscalefactor); dot((0.78,-1.68),dotstyle); label("$A$", (0.86,-1.48), NE * labelscalefactor); dot((3.204081378819364,3.962943537579509),linewidth(4pt) + dotstyle); label("$C$", (3.28,4.12), NE * labelscalefactor); dot((-1.29430082475954,1.790464841424615),linewidth(4pt) + dotstyle); label("$B$", (-1.22,1.96), NE * labelscalefactor); dot((4.808930448090646,1.5511224385677451),linewidth(4pt) + dotstyle); label("$D$", (4.88,1.72), NE * labelscalefactor); dot((-0.3565251960972488,1.509518912147643),linewidth(4pt) + dotstyle); label("$I_{B}$", (-0.28,1.66), NE * labelscalefactor); dot((0.9625586859152774,-0.6655982628527022),linewidth(4pt) + dotstyle); label("$I_{A}$", (1.04,-0.5), NE * labelscalefactor); dot((4.108197259988662,1.6597951557426085),linewidth(4pt) + dotstyle); label("$I_{D}$", (4.18,1.82), NE * labelscalefactor); dot((3.02450048712704,3.306095524987752),linewidth(4pt) + dotstyle); label("$I_{C}$", (3.1,3.46), NE * labelscalefactor); dot((1.1441169545779593,0.34324457692514415),linewidth(4pt) + dotstyle); label("$J_{A}$", (1.22,0.5), NE * labelscalefactor); dot((0.5864177373916837,1.2270249225329872),linewidth(4pt) + dotstyle); label("$J_{B}$", (0.66,1.38), NE * labelscalefactor); dot((3.403602911632138,1.7690666782594093),linewidth(4pt) + dotstyle); label("$J_{D}$", (3.48,1.92), NE * labelscalefactor); dot((2.845903694445863,2.6528470238672526),linewidth(4pt) + dotstyle); label("$J_{C}$", (2.92,2.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Firstly notice that $$\frac{BL}{CL}=\frac{BL}{LK}\cdot\frac{CL}{LM}=\frac{\sin\angle ABC}{\sin\angle BCD}=\frac{AC}{BD}$$and similar ratio holds. Let $I_A$ be the incenter of $\triangle AKN$, $J_A$ be the incenter of $\triangle ABD$, and define $I_B,I_C,I_D,J_B,J_C,J_D$ symmetrically. Then a homothety at $C$ with ratio $\frac{BD}{BD+AC}$ will send $J_C$ to $I_C$. \newline\newline Now we can basically ignore the rhombus. \newline\newline Lemma. Suppose $J_CJ_A$ meet $BD$ at $J$, $I_BJ_B$ meet $AC$ at $O$, then $$\frac{BJ}{JD}=\frac{J_BO}{J_BO}$$\newproof Just some trigonometry. $$\frac{BJ}{JD}=\frac{BJ}{JI_A}\cdot\frac{JI_A}{JD}=\frac{\sin\angle BI_AI_C}{\sin\angle DI_AI_C}\cdot\frac{\sin\frac{\angle ADB}{2}}{\sin\frac{\angle ABD}{2}}=\frac{\sin\frac{\angle ADB}{2}\sin\frac{\angle ABC}{2}\sin\frac{\angle BDC}{2}}{\sin\frac{\angle ABD}{2}\sin\frac{\angle ADC}{2}\sin\frac{\angle DBC}{2}}$$while $$\frac{I_BO}{I_DO}=\frac{I_BO}{AO}\cdot\frac{AO}{I_DO}=\frac{\sin\frac{\angle BDC}{2}}{\sin\frac{\angle DBC}{2}}\cdot\frac{AI_B}{AI_D}$$Notice that $AI_B=4R\sin\frac{\angle ABC}{2}\sin\frac{\angle ADB}{2}$ and $AI_D=4R\sin\frac{\angle ABD}{2}\sin\frac{\angle ADC}{2}$ so we are done.$\square$ [asy][asy] size(10cm); 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 = -5.466557332771997, xmax = 15.367236717516683, ymin = -4.39878782481106, ymax = 7.839898366165694; /* image dimensions */ pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen zzttff = rgb(0.6,0.2,1); /* draw figures */ draw((-1.1070314065655054,1.16505256005646)--(0.582716948645905,-1.6894603474192578), linewidth(0.8) + fuqqzz); draw((0.582716948645905,-1.6894603474192578)--(6.991047795903257,-0.17268323740150626), linewidth(0.8) + fuqqzz); draw((6.991047795903257,-0.17268323740150626)--(0.5414271913783777,5.174123280827074), linewidth(0.8) + fuqqzz); draw((0.5414271913783777,5.174123280827074)--(-1.1070314065655054,1.16505256005646), linewidth(0.8) + fuqqzz); draw(circle((3.150433553807177,1.7579024603968663), 4.298543810772993), linewidth(0.8) + blue); draw((0.5414271913783777,5.174123280827074)--(0.582716948645905,-1.6894603474192578), linewidth(0.8) + qqwuqq); draw((-1.1070314065655054,1.16505256005646)--(-0.22571641538166162,1.2287834470055459), linewidth(0.8)); draw((2.5854384685702647,0.8583280334438508)--(6.991047795903257,-0.17268323740150626), linewidth(0.8)); draw((0.5414271913783777,5.174123280827074)--(1.217670301646677,2.460781095442067), linewidth(0.8)); draw((0.582716948645905,-1.6894603474192578)--(1.142051751541926,-0.37366961499267026), linewidth(0.8)); draw((-1.1070314065655054,1.16505256005646)--(6.991047795903257,-0.17268323740150626), linewidth(0.8) + ffvvqq); draw((-0.6270749150749081,1.1997598476955584)--(4.591791211393503,0.388796548137482), linewidth(0.8) + ffvvqq); draw((0.8374428747475465,-1.090237822487931)--(0.849394175805358,3.93844368888269), linewidth(0.8) + qqwuqq); draw((1.217670301646677,2.460781095442067)--(1.142051751541926,-0.37366961499267026), linewidth(0.8) + qqwuqq); draw((-0.22571641538166162,1.2287834470055459)--(2.5854384685702647,0.8583280334438508), linewidth(0.8) + ffvvqq); draw((0.5657889330357238,1.1244784715457623)--(1.1730539196655738,0.7884012967262525), linewidth(0.8) + linetype("4 4") + zzttff); /* dots and labels */ dot((0.582716948645905,-1.6894603474192578),dotstyle); label("$A$", (-0.23475254884300154,-1.9230230609875179), NE * labelscalefactor); dot((0.5414271913783777,5.174123280827074),linewidth(4pt) + dotstyle); label("$C$", (0.6294294913595558,5.364133602342152), NE * labelscalefactor); dot((-1.1070314065655054,1.16505256005646),linewidth(4pt) + dotstyle); label("$B$", (-1.0055095036182555,1.3468549289681033), NE * labelscalefactor); dot((6.991047795903257,-0.17268323740150626),linewidth(4pt) + dotstyle); label("$D$", (7.0757603858434965,0.015547461629028977), NE * labelscalefactor); dot((-0.6270749150749081,1.1997598476955584),linewidth(4pt) + dotstyle); label("$I_{B}$", (-0.6551654332658673,1.6738427279636654), NE * labelscalefactor); dot((0.8374428747475465,-1.090237822487931),linewidth(4pt) + dotstyle); label("$I_{A}$", (0.9330610189982923,-0.8953471212871797), NE * labelscalefactor); dot((4.591791211393503,0.388796548137482),linewidth(4pt) + dotstyle); label("$I_{D}$", (4.693420707447259,0.5760979741928497), NE * labelscalefactor); dot((0.849394175805358,3.93844368888269),linewidth(4pt) + dotstyle); label("$I_{C}$", (0.9330610189982923,4.126251220430381), NE * labelscalefactor); dot((1.142051751541926,-0.37366961499267026),linewidth(4pt) + dotstyle); label("$J_{A}$", (1.2366925466370287,-0.1946589805824038), NE * labelscalefactor); dot((-0.22571641538166162,1.2287834470055459),linewidth(4pt) + dotstyle); label("$J_{B}$", (-0.3515339056271309,1.720555270677317), NE * labelscalefactor); dot((2.5854384685702647,0.8583280334438508),linewidth(4pt) + dotstyle); label("$J_{D}$", (2.684781370760233,1.043223401329367), NE * labelscalefactor); dot((1.217670301646677,2.460781095442067),linewidth(4pt) + dotstyle); label("$J_{C}$", (1.3067613607075064,2.6548061249503516), NE * labelscalefactor); dot((0.8423426810011356,0.9714260078794937),linewidth(4pt) + dotstyle); label("$I$", (0.9330610189982923,1.1600047581134965), NE * labelscalefactor); dot((1.1730539196655738,0.7884012967262525),linewidth(4pt) + dotstyle); label("$J$", (1.2600488179938545,0.9731545872588895), NE * labelscalefactor); dot((0.5657889330357238,1.1244784715457623),linewidth(4pt) + dotstyle); label("$O$", (0.2090166069366901,1.5337050998227102), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] We now return to the original problem. Let $X_1$ be the internal homothetic center of $\omega_1$ and $\omega_3$, and let $X_2$ be the internal homothetic center of $\omega_2$ and $\omega_4$. \newline\newline The main idea here is spiral similarity. Notice that $$\frac{CI_C}{I_CJ_C}=\frac{AI_A}{I_AJ_A}$$so let $X$ be the center of spiral sim. sending $CJ_C$ to $AJ_A$, then it sends $I_C$ to $I_A$ as well. Meanwhile, notice that $J$ is the internal homothetic center of the incircle of $\triangle CBD$ and $\triangle ABD$. Let the inradius of $\triangle ABD$ be $r_A$, and define $r_B,r_C,r_D$ symmetrically. Then from the lemma and homothety, $$\frac{CO}{OA}=\frac{I_CX_1}{X_1I_A}=\frac{J_CJ}{J_CJA}=\frac{r_C}{r_A}$$Therefore, $$\angle XJO=\angle XJ_CC=\angle XJ_CI_C=\angle XJX_1$$hence $O,J,X_1$ are collinear, by symmetry $O,J,X_2$ are collinear. Moreover, $$\frac{OX_1}{X_1J}=\frac{CI_C}{I_CJ_C}=\frac{BD}{AC}$$while $$\frac{OX_2}{X_2J}=\frac{I_BJ_B}{BI_B}=\frac{BD}{AC}$$as well, so $X_1=X_2$ as desired.

Lemma: let $(C_1,O_1),...,(C_4,O_4)$ be circles such that $ex(C_1,C_2)=ex(C_3,C_4)=E$ and $ex(C_1,C_4)=ex(C_2,C_3)=F$ ($in(,)$ and $ex(,)$ are in- and exsimilicenter respectively). Then $in(C_1,C_3)=in(C_2,C_4)$ Proof: let $G=O_1O_3\cap O_2O_4$, $H=ex(C_1,C_3)$, $I=ex(C_2,C_4)$. By Monge theorem on $C_1,C_2,C_3$ and $C_1,C_4,C_3$ we have that $H,I\in EF$. Therefore, $(O_1O_3,GH)=-1$ and thus $G=in(C_1,C_3)$, and similarly we must have $G=in(C_2,C_4)$, and we are done. Going back to the problem, let $E=AD\cap BC$, $F=AB\cap CD$, $G=AC\cap BD$. It easy to see by angle chasing that the bisector of $\angle BEA$ is parallel to the bisector of $\angle BGA$, and thus also parallel to the angle bisector of $\angle NML$ (since $MN||AG$ and $LM||BG$), i.e. the line $KM$. Similarly the bisector of $\angle BFC$ and $LN$ are parallel. Now the two bisectors are also heights of the triangles $ELN$ and $FKM$, and thus both of these triangles are isosceles with vertices $E,F$ (note that we must also have $E\in KM$, $F\in LN$). Since $FKM$ is isosceles and $N$ belongs on its height, it follows that $MKXY$ is an isosceles trapezoid, where $X=AB\cap MN$ and $Y=CD\cap KN$. Equivalently, $FXNY$ must be a kite, and thus has an incircle $\omega$. Now, by Monge's Theorem on $\omega,\omega_A,\omega_D$, it follows that $K,M$ and $ex(\omega_A,\omega_D)$ are aligned, but since we already know that $AD$ is a common tangent of $\omega_A$ and $\omega_D$, it follows that $ex(\omega_A,\omega_D)=AD\cap KM=E$. Similarly $ex(\omega_B,\omega_C)=E$ and so the two exsimilicenters coincide in $E$. Analogously, $ex(\omega_A,\omega_B)\equiv ex(\omega_3,\omega_4)=F$. Thus, by our lemma, $in(\omega_A,\omega_C)\equiv in(\omega_B,\omega_D)$, and thus we are done.

We will prove this problem for any parallelogram $KLMN$ with $\overline{KL} \parallel \overline{AC}$ and $\overline{LM} \parallel \overline{BD}$. Let $\omega_A$ be tangent to $\overline{NK}$ at $E$, and define $F$, $G$, and $H$ similarly. Let $\overline{EG}$ and $\overline{FH}$ intersect at $X$. We claim that $X$ is the insimilicenter of $\omega_A$ and $\omega_C$, and similarly, the insimilicenter of $\omega_B$ and $\omega_D$. Let $r_A$, $r_B$, $r_C$, and $r_D$ denote the radii of $\omega_A$, $\omega_B$, $\omega_C$, and $\omega_D$, respectively. It is sufficient to prove $\frac{r_A}{r_C}=\frac{EX}{XG}$, since then we can take a homothety centered at $X$ that takes $E$ to $G$, which must also take $\omega_A$ to $\omega_C$ (since $E$ and $G$ are the corresponding "bottom" points of the incircles). Then, do algebra, which has to work out. Don't wanna write out that many details

Let $E=\overline{AB}\cap\overline{CD},$ $F=\overline{BC}\cap\overline{AD},$ $N_1=\overline{NK}\cap\overline{BC},$ and $L_1=\overline{LK}\cap\overline{AD}.$ Also, denote the insimilicenter of two circles $\Omega$ and $\Gamma$ as $i(\Omega,\Gamma)$ and the exsimilicenter as $e(\Omega,\Gamma.)$ Claim: $E$ lies on $\overline{LN}.$ Proof. Notice $$\angle EKL=\angle EAC=\angle BDC=\angle LME$$so $\angle EKM=\angle KME$ and $EK=EM.$ Hence, $\triangle ELK\cong\triangle ELM$ and $\delta(L,\overline{AE})=\delta(L,\overline{DE}).$ Similarly, $\triangle ENK\cong\triangle ENM$ and $\overline{LN}$ is the angle bisector of $\angle AED.$ $\blacksquare$ Claim: $E$ is the exsimilicenter of $\omega_A$ and $\omega_B.$ Proof. We claim $FN_1KL_1$ has an incircle. Notice $\overline{FK}$ bisects $\angle N_1FL_1$ and $$\measuredangle FL_1K=\measuredangle DL_1K=\measuredangle DAC=\measuredangle DBC=\measuredangle NN_1C=\measuredangle KN_1F.$$Hence, by symmetry, the angle bisectors of $\angle KN_1F$ and $\angle KL_1F$ concur on $\overline{FK},$ and $FN_1KL_1$ has an incircle. By Monge on the incircle, $\omega_A,$ and $\omega_B,$ we have $L,N,$ and $e(\omega_A,\omega_B)$ are collinear. But $E=\overline{BC}\cap\overline{LN}$ so $E=e(\omega_A,\omega_B).$ $\blacksquare$ By 2i1e Monge on $\omega_A,\omega_B,\omega_C,$ we have $e(\omega_A,\omega_B),i(\omega_B,\omega_C),i(\omega_A,\omega_C)$ are collinear. By 2i1e Monge on $\omega_A,\omega_C,\omega_D,$ we have $e(\omega_C,\omega_D),i(\omega_A,\omega_D),i(\omega_A,\omega_C)$ are collinear. Similarly, By 2i1e Monge on $\omega_A,\omega_B,\omega_D,$ we have $e(\omega_A,\omega_B),i(\omega_A,\omega_D),i(\omega_B,\omega_D)$ are collinear. By 2i1e Monge on $\omega_B,\omega_C,\omega_D,$ we have $e(\omega_C,\omega_D),i(\omega_B,\omega_D),i(\omega_B,\omega_C)$ are collinear. Hence, $i(\omega_A,\omega_C)$ and $i(\omega_B,\omega_D)$ both lie on the line connecting $i(\omega_B,\omega_C)$ and $i(\omega_A,\omega_D).$ Similarly, they lie on the line connecting $i(\omega_A,\omega_B)$ and $i(\omega_C,\omega_D)$ so they are the same point. $\square$ Remarks: (Motivation) I wanted to use Monge because of the tangent circles, so I drew the external tangents of $\omega_B$ and $\omega_C.$ I realized that the tangents concurred with $\overline{BC},$ $\overline{AD},$ and $\overline{LN}.$ The collinearity finish with Monge was natural after drawing internal tangents. The second claim was the most difficult to prove. I wanted to use Monge to prove $L,N,$ and $e(\omega_B,\omega_C)$ are collinear. The extension of the sides and the incircle construction were natural after this.

Let $AB$ and $CD$ meet at $S$ and $AD$ and $BC$ meet at $T$. Claim $: T,M,K$ and $S,N,L$ are collinear. Proof $:$ we'll prove $T,K,M$ all lie on angle bisector of $\angle DTC$. First Note that $\frac{DM}{DC} = \frac{\frac{DM}{DC}}{\frac{CM}{DC}} = \frac{\frac{NM}{AC}}{\frac{ML}{BD}} = \frac{BD}{AC} = \frac{TD}{TC} \implies TM$ is angle bisector of $\angle DTC$ Now Note that $TDC$ and $TBA$ are similar and $\frac{DM}{MC} = \frac{DN}{NA} = \frac{BK}{KA}$ so $T,M,K$ are collinear. we'll use exact same approach for proving $S,N,L$ are collinear. Claim $: S$ is exsimilicenter of $\omega_d,\omega_c$ meet. Proof $:$ Let $MN$ and $ML$ meet $TC$ and $TD$ at $X,Y$ Note that $TXMY$ is kite cause $TNL$ is isosceles and $\angle TLM = \angle CBD = \angle CAD = \angle TNM$ so let it's incircle be $\omega_t$. Now with Monge Theorem on $\omega_t,\omega_d,\omega_c$ we have exsimilicenter of $\omega_d,\omega_c$ lies on $LN$ so $S$ is exsimilicenter. Now with Monge Theorem on $\omega_a,\omega_d,\omega_c$ and $\omega_a,\omega_b,\omega_c$ we have insimilicenter of $\omega_a,\omega_c$ is intersection of $TP,SQ$ where $Q,P$ are insimilicenters of $\omega_a,\omega_d$ and $\omega_c,\omega_d$. we have same approach for insimilicenter of $\omega_b,\omega_d$ so their insimilicenters is same.

Let $X=AB\cap CD,Y=AD\cap BC;P,Q$ are insimilicenters of pairs $(\omega_A,\omega_B),(\omega_A,\omega_D)$ respectively. Note that $\angle BKL=\angle BAC=\angle BDC=\angle LMC,$ so by symmetry $X\in LN.$ Analogously $Y\in KM.$ Furthermore, complete quadrilateral $AB,CD,KL,LM$ has incircle, so by Monge $Y=KM\cap BC$ is the exsimilicenter of $\omega_B,\omega_C$ and by the same reason of $\omega_A,\omega_D.$ Similarly we proceed for $Y=LN\cap CD,$ so by Monge on $\omega_A,\omega_B,\omega_C$ and $\omega_A,\omega_B,\omega_D$ point $XQ\cap YP$ is the incimilicenter of $\omega_B,\omega_D.$ Analogously it is the incimilicenter of $\omega_A,\omega_C$ as desired.

$ $

Attachments:

Solved with Carried by OronSH and ihatemath123 Let $AD\cap BC=X$ and $AB\cap CD=Y$. Notice that if we intersect the angle bisectors of $\angle BXA$ and $\angle AYD$ with the cyclic quadrilateral that we get a rhombus, which is $KLMN$ meaning $KMX$, $LNY$ are collinear. Claim: $X$ is the exsimilicenter of $\omega_A,\omega_D$ and if $\omega_B,\omega_C$. Similar relations hold for $Y$. Proof: We show that there exists a circle tangent to $AK,KN,MD,MN$ which will imply the result by monge of said circle, $\omega_A,\omega_D$ and using symmetry. This works as $X=KM\cap AD.$ Now, we prove this. It suffices to show lines $MI_D,KI_A,LN$ concur as they are all angle bisectors. But this is true by symmetry as $\angle KMD=\angle MKA$ and subtracting $\angle KMN=\angle MKN$ gives $\angle AKN=\angle NMD$ so the bisectors are symmetric with respect to the perpendicular bisector of $MK,$ which is $LN$ proving the claim. $\square$ Now, Monge gives the exsimilicenters of $\omega_A,\omega_C$ and $\omega_B,\omega_D$ lie on $XY.$ Using the fact that centers of circles and their similicenters produce harmonic bundles finishes and we can project the bundles formed by $\omega_A,\omega_C$ and the one formed by $\omega_B,\omega_D$ onto each other through $X,Y$ to finish.

agram

Let $AB$ and $CD$ intersect at $Q$, while $BC$ and $AD$ intersect at $P$. Without loss of generality, let $Q$ be on the extensions of $BA$ and $DA$. Note that \[\angle NKQ=\angle DBQ=\angle QCA=\angle QMN\]so by symmetry, $QN$ bisects $\angle KQM$. Similarly, $QL$ bisects $KPM$. In particular, $Q$, $N$, $L$ are collinear. Similarly, $P$, $K$, $M$ are collinear. Since $KNMQ$ is a kite, if $KN$ intersects $MQ$ at $F$ and $MN$ intersects $KQ$ at $G$, then $FNGQ$ is a kite and is thus tangential. By Monge, the exsimilicenters taken pairwise on the incircle of $FNGQ$, $\omega_A$ and $\omega_D$ gives that the exsimilicenter of $\omega_A$ and $\omega_D$ lies on $MK$. Since it also lies on $AD$, it is $P$. Similarly, $P$ is the exsimilicenter of $\omega_B$ and $\omega_C$. Additionally, $Q$ is the exsimilicenter of $\omega_A$ and $\omega_B$ as well as $\omega_C$ and $\omega_D$. Let $X$ be the insimilicenter of $\omega_A$ and $\omega_B$ and let $Y$ by the insimilicenter of $\omega_C$ and $\omega_D$. By Monge on $\omega_A$, $ omega_B$, and $\omega_C$, the insimilicenter of $\omega_A$ and $\omega_C$ lies on $PX$. By Monge on $\omega_A$, $\omega_B$, and $\omega_D$, the insimilicenter of $\omega_B$ and $\omega_D$ also lies on $PX$. Similarly, both insimilicenters lie on $QY$ so they are the same point, as desired.

Let $BC \cap DA = X$ and $AB \cap CD = Y$. Our problem rests in the following claim: Claim: $KM$ bisects $\angle XAB$, and thus $\triangle XLN$ is isosceles, and similarily for $\triangle YKM$. We show $XK$ and $XM$ are both bisectors through Angle Bisector Theorem. Using the similarity ratios, we find \[\frac{AK}{BK} = \frac{\frac{AK}{AB}}{\frac{BK}{AB}} = \frac{\frac{NK}{BD}}{\frac{KL}{AC}} = \frac{AC}{BD} = \frac{XA}{XB} = \frac{XC}{XD},\] which analogously is equal to $\frac{CM}{DM}$. ${\color{blue} \Box}$ Therefore we find that the incircles of $\triangle LMX$ and $\triangle NMX$ are congruent and reflections over $XM$. We use Monge's on: $\omega_C$ and the incircles of $\triangle LMX$ and $\triangle NMX$: Exsimilicenter of $\omega_C$ and the incircle of $\triangle NMX$ lies on $NY$. $\omega_C$, $\omega_D$, and the incircle of $\triangle NMX$: Exsimilicenter of $\omega_C$ and $\omega_D$ is $Y$. Similarily, the exsimilicenter of $\omega_A$ and $\omega_B$ is $Y$. Similarily, the exsimilcenter of $\omega_B$, $\omega_C$ and $\omega_D$, $\omega_A$ is $X$. Hence we can finish with triples of Monge's on $\omega_A$, $\omega_B$, $\omega_C$, and $\omega_D$. $\blacksquare$

Let $\mathcal{I}(\omega_X, \omega_Y)$ denote the insimilicenter of $\omega_X$ and $\omega_Y$, and let $\mathcal{E}$ denote the exsimilicenter similarly. Let $P = \overline{AD} \cap \overline{BC}$ and $Q = \overline{AB} \cap \overline{CD}$. See that $\angle PNK = \angle BDP = \angle PCA = \angle PLK$ which implies that $PM$ is the angle bisector of $\angle NML$, and coincides with $KM$. By symmetry, we also have $QL$ being the angle bisector of $\angle{KLM}$. Then let $X = \overline{MN} \cap \overline{BC}$ and $Y = \overline{LM} \cap \overline{AD}$. It is clear that $PXMY$ is a kite, with incircle $\omega_P$. Since $NM$ and $ND$ are both tangents to $\omega_D$, $\mathcal{E}(\omega_D, \omega_P) = N$ and similarly, $\mathcal{E}(\omega_C, \omega_P) = L$. Then by Monge's theorem, we have $N$, $L$, and $\mathcal{E}(\omega_C, \omega_D)$ are collinear. Notice that $\mathcal{E}(\omega_C, \omega_D)$ also lies on $\overline{CD}$, and $\overline{CD} \cap \overline{NL} = Q$, so $\mathcal{E}(\omega_C, \omega_D) = Q$. By symmetry, we also have $\mathcal{E}(\omega_A, \omega_B) = Q$ and $\mathcal{E}(\omega_A, \omega_D) = \mathcal{E}(\omega_B, \omega_C) = P$. Then doing Monge's on $(\omega_B, \omega_C, \omega_D)$ gives $Q$, $\mathcal{I}(\omega_B, \omega_C)$, and $\mathcal{I}(\omega_B, \omega_D)$ collinear. Similarly, Monge's on $(\omega_A, \omega_B, \omega_C)$ gives $Q$, $\mathcal{I}(\omega_B, \omega_C)$, and $\mathcal{I}(\omega_A, \omega_C)$ collinear. Repeating the process for $(\omega_A, \omega_B, \omega_D)$ and $(\omega_A, \omega_C, \omega_D)$ gives us our concurrency.