This is really a perfect corollary for the Monge-D'Alembert circle theorem and its 1-2 variant. Theorem 1 (Monge-D'Alembert). The exsimilicenters of three given circles in general position are collinear. Proof of Theorem 1. This is a simple consequence of Menelaus' theorem, since the exsimilicenter $E$ of two circles $\mathcal{C}_{1}(O_{1},r_{1})$ and $\mathcal{C}_{2}(O_{2},r_{2})$ satisfies $EO_{1}/EO_{2} = r_{1}/r_{2}$ (the segments are not oriented). The 1-2 variant of Theorem 1. Let there be three circles in general position. The exsimilicenter of two of them lies on the line determined by the insimilicenters of the other two pairs. Consider $\Gamma(O)$ the circle tangent to the lines $AB$, $BC$, $AD$ of the quadrilateral $ABCD$ (this circle obviously exist, according to Apollonius) and denote by $\rho_{1}$, $\rho_{2}$, $\rho_{p}$ the incircles of triangles $APD$, $BPC$ and $CPD$ respectively. Since $A$ is the exsimilicenter of $\rho_{1}$ and $\Gamma$ and $K$ is the insimilicenter of $\rho_{1}$ and $\rho_{p}$, according to the 1-2 variant of the Monge-D'Alembert theorem, the line $AK$ intersects the line $OI$ at the insimilicenter of $\Gamma$ and $\rho_{p}$. Similarly, we obtain that the line $BK$ intersects $OI$ at the same insimilicenter $F$ of $\Gamma$ and $\rho_{p}$. Thus, it remains to prove that $E$ lies on the line $OI$. It is easy to see now that the quadrilaterals $APCD$ and $PBCD$ are circumscribed. Denote by $\omega_{a}$ and $\omega_{b}$ their incircles. We proceed similarly: $A$ is the exsimilicenter of $\omega_{a}$ and $\Gamma$, $C$ is the exsimilicenter of $\omega_{a}$ and $\rho_{p}$, and thus, according this time to the original Monge-D'Alembert theorem, the line $AC$ intersects the line $OI$ at the exsimilicenter of the circles $\rho_{p}$ and $\Gamma$. Analogously, $B$ is the exsimilcenter of $\omega_{b}$ and $\Gamma$, $D$ is the exsimilcenter of $\rho_{p}$ and $\omega_{b}$, and thus $BD$ intersects $OI$ at the exsimilicenter of $\Gamma$ and $\rho_{p}$. In this case, we conclude that $E$, $F$ are the insimilicenter and exsimilicenter of the circles $\Gamma$ and $\rho_{p}$, respectively. This proves the problem and moreover it shows that the cross-ratio $(O, E, I, F)$ is harmonic.

April wrote: Point $P$ lies on side $AB$ of a convex quadrilateral $ABCD$. Let $\omega$ be the incircle of triangle $CPD$, and let $I$ be its incenter. Suppose that $\omega$ is tangent to the incircles of triangles $APD$ and $BPC$ at points $K$ and $L$, respectively. Let lines $AC$ and $BD$ meet at $E$, and let lines $AK$ and $BL$ meet at $F$. Prove that points $E$, $I$, and $F$ are collinear. Author: Waldemar Pompe, Poland Double Desargue! Lemma 1: If the incircles of $WXY$ and $WZY$ in convex quadrilateral $WXYZ$ are tangent, then $WXYZ$ is circumscriptible. Suppose the incircles are tangent to each other at $T$. Then we have $$WX+WY-XY=2WT=WZ+WY-YZ$$ which implies that $WX+YZ=WZ+XY$, so the result follows. Lemma 2: If a circle $\Gamma$ is externally tangent to two circles $\Gamma_1$ and $\Gamma_2$ at $M$ and $N$, respectively, then $MN$ passes through the exsimilicenter of $\Gamma_1$ and $\Gamma_2$. This is a variant of the Monge circle theorem, but can be proven using simple homothety. Proof of Problem: Let $I_1$ be the incenter of $APD$ $I_2$ be the incenter of $BPC$ $I_1'$ be the incenter of $APCD$ (this exists by Lemma 1) $I_2'$ be the incenter of $BPDC$ ( ) $G$ be the intersection of $AI_1$ and $AI_2$. I claim that $E,F,G,I$ lie on a line. By Desargue on $AI_1K$ and $BI_2L$ and Lemma 2, $FIG$ is a line. (as $AK\cap BL=F$, $I_1K\cap I_2L=I$, and $AI_1\cap BI_2=G$. Note that $AB,CD,I_1'I_2'$ concur on the exsimilicenter of the incircles of $APCD$ and $BPDC$. Thus by Desargue on $AI_1'C$ and $BI_2'D$, $EIG$ is a line, as $AI_1'\cap BI_2'=G$, $CI_1'\cap DI_2'=I$, and $AC\cap BD=E$. This completes the proof.

Only a single Desargue is needed.

Truly a beautiful problem

This is very very beautiful! It can be done using Monge d'Alembert theorem multiple times. In my solution I also used three constructed circles.

Let $BC$ and $AD$ meet at $G$. Let $\omega$ be the incircle of $PCD$. Let $\omega_1$ and $\omega_2$ be the incircles of $PAD$ and $PBC$. Let $\Gamma$ be the incircle of $GAB$. Note that $$\frac{PD+PC-CD}{2}=\frac{PA+PK-AD}{2}$$ so $PA+CD=PC+AD$ and $PCDA$ is circumscribed. Similarly, so is $PDCB$. Let $\Gamma_1$ and $\Gamma_2$ be the incircles of these quadrilaterals, respectively. Let $Se(X,Y)$ denote the exsimilicenter of two circles $X,Y$, and let $Si(X,Y)$ denote their insimilicenter. By Monge on $\Gamma,\omega_1,\omega,$ $Si(\Gamma,\omega)$ lies on $AK$ and similarly $BL$. By Monge on $\Gamma_1,\omega_1,\omega$, $Se(\omega_1,\omega)$ lies on $AC$. By Monge on $\Gamma, \omega_1,\omega$, $Se(\gamma,\omega)$ is collinear with $A$ and $Se(\omega_1,\omega)$. So it also lies on $AC$, and similarly $BD$. Thus the three points are collinear on the line through the centers of $\omega$ and $\Gamma$.

Very beautiful! We begin by noticing that the quadrilaterals $PADC$ and $PBCD$ are circumscribed. Indeed, we have $$PA-AD=PK-KD= PC-CD$$ and by the converse of Pithot's Theorem, the claim holds. A similar argument works for $PBCD$. Let us denote by $\omega$ the incircle if triangle $PDC$ with center $I$ and by $\Omega$ the circle tangent to the lines $AB,BC,AD$ with center $O$ and let circles $\omega_1$ and $\omega_2$ denote the incircles of triangles $PAD$ and $PBC$, respectively. Denote by $\Omega_1$ and $\Omega_2$ the inscribed circles in the quadrilaterals $PADC$ and $PBCD$, respectively. Let $E',F'$ be the exsimilicenter and insimilicenter, respectively, of the circles $\omega$ and $\Omega$. The points $O,I,E',F'$ are collinear and $(O,I,E',F')=-1$. Next, we show that $A,C,E'$ are collinear and $A,K,F'$ are collinear. Similar statements shall hold for $B$ and we will get $E=E'$ and $F=F'$ proving the result. Notice that $A$ is the exsimilicenter for circles $\omega_1$ and $\Omega$ and $K$ is the insimilicenter for circles $\omega_1$ and $\omega$. Applying Monge D'Alembert's Theorem, we see that points $A,K,F'$ are collinear. Notice that $A$ is the exsimilicenter for circles $\omega_1$ and $\Omega_1$ and $C$ is the exsimilicenter for circles $\Omega_1$ and $\omega$ thus, by Monge D'Alembert's Theorem, line $AC$ passes through the exsimilicenter of circles $\omega_1$ and $\omega$. Also, note that $A$ is the exsimilicenter of circles $\omega_1$ and $\Omega$ and so, by Monge D'Alembert's Theorem, line $AE'$ passes through the exsimilicenter of circles $\omega_1$ and $\omega$. This proves that $A,E',C$ are collinear and the conclusion holds.

Let $\omega_A$ and $\omega_B$ be the incircles of $\triangle APD$ and $\triangle BPC$, and let $I_A$ and $I_B$ be their centers. Let lines $AI_A$ and $BI_B$ meet at $J$. Then let $\omega_J$ be the circle centered at $J$ tangent to lines $AB$, $AD$, and $BC$. Now, note $$AP - AD = KP - KD = CP - CD,$$ so $APCD$ has an incircle $\gamma_D$. Similarly define $\gamma_C$. We are now ready to solve the problem. By Monge on $\gamma_C, \omega, \omega_J$, the exsimilicenter for $\omega$ and $\omega_J$ belongs to $AC$; similarly it belongs to $BD$ and it is $E$. By Monge on $\omega_A, \omega, \omega_J$, the insimilicenter for $\omega$ and $\omega_J$ belongs to $AK$; similarly it belongs to $BL$ and it is $F$. Thus $E$, $F$, $I$, $J$ are collinear.

Solution from Twitch Solves ISL: We begin with the following claim. Claim: The quadrilaterals $APCD$ and $PBCD$ both have an incircle. Proof. Follows by Pitot theorem since $AP+CD = PC+DA$ (see SL 2017 G7 for another example of this lemma). $\blacksquare$ Let $I$ be the center of $\omega$ (the incenter of $\triangle CPD$). Moreover, if we let $G = \overline{AD} \cap \overline{BC}$, we will denote by $J$ the incenter of $\triangle GBA$, and by $(J)$ the circumcircle. [asy][asy] size(12cm); pair D = (11.14,-6.04); pair C = (2.,-6.); pair P = (4.444652508480686,1.8445570362182018); pair I = (5.51778691510771,-3.4234688887952602); pair K = (7.493465232949166,-1.745777572891788); pair L = (3.0432615410670767,-2.652315765041075); pair T = (5.5064439102990566,-6.015345487572424); pair A = (15.364272954643416,5.744248447444408); pair B = (-3.880008641661257,-1.1284043179454462); pair F = (5.883808240537381,-3.2775609912250143); pair J = (6.629240970901842,-2.980407478817377); pair G = (9.058135630380526,-11.847675599776444); filldraw(B--A--G--cycle, invisible, deepcyan); filldraw(incircle(B, A, G), invisible, deepcyan); filldraw(P--C--D--cycle, invisible, red); filldraw(incircle(P, C, D), invisible, red); draw(B--F--A, dotted); draw(A--C, dotted); draw(B--D, dotted); pair E = extension(B, D, A, C); draw(incircle(B, P, C), deepgreen); draw(incircle(A, P, D), deepgreen); pair I_1 = extension(D, I, B, incenter(B, P, C)); filldraw(CP(I_1, foot(I_1, C, D)), invisible, orange+dashed); pair I_2 = extension(C, I, A, incenter(D, P, A)); filldraw(CP(I_2, foot(I_2, C, D)), invisible, orange+dashed); dot("$D$", D, dir(D)); dot("$C$", C, dir(225)); dot("$P$", P, dir(120)); dot("$I$", I, dir(I)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$T$", T, dir(T)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$F$", F, dir(F)); dot("$J$", J, dir(J)); dot("$G$", G, dir(G)); dot("$E$", E, dir(E)); /* TSQ Source: !size(12cm); D = (11.14,-6.04) .C = (2.,-6.) R225 .P = (4.444652508480686,1.8445570362182018) R120 I = (5.51778691510771,-3.4234688887952602) K = (7.493465232949166,-1.745777572891788) L = (3.0432615410670767,-2.652315765041075) T = (5.5064439102990566,-6.015345487572424) A = (15.364272954643416,5.744248447444408) B = (-3.880008641661257,-1.1284043179454462) F = (5.883808240537381,-3.2775609912250143) J = (6.629240970901842,-2.980407478817377) G = (9.058135630380526,-11.847675599776444) B--A--G--cycle 0.1 lightcyan / deepcyan incircle B A G 0.1 lightcyan / deepcyan P--C--D--cycle 0.1 lightred / red incircle P C D 0.1 lightred / red B--F--A dotted A--C dotted B--D dotted E = extension B D A C incircle B P C deepgreen incircle A P D deepgreen I_1 := extension D I B incenter B P C CP I_1 foot I_1 C D 0.1 yellow / orange dashed I_2 := extension C I A incenter D P A CP I_2 foot I_2 C D 0.1 yellow / orange dashed */ [/asy][/asy] Claim: The point $F$ is the insimilicenter of $(I)$ and $(J)$. Proof. By Monge d'Alembert theorem on $(I)$, $(J)$, and the incircle of $\triangle APB$, we get that $\overline{BL}$ passes through said insimilicenter. By symmetry, so does $\overline{CK}$, as needed. $\blacksquare$ Claim: The point $E$ is the exsimilicenter of $(I)$ and $(J)$. Proof. By Monge d'Alembert theorem on $(I)$, $(J)$, and the incircle of $BPDC$, we get that $\overline{BD}$ passes through said exsimilicenter. By symmetry, so does $\overline{CA}$, as needed. $\blacksquare$ The previous two claims imply all four points $E$, $F$, $I$, $J$ are collinear.

First we claim that the Quadrilaterals $APCD$ and $BPDC$ are circumscribed . Note that $$PK= \frac {PD+PC-DC}{2} = \frac {PA+PD-AD}{2} \implies AD + PC=PA+CD$$ Hence by converse of Pitot's theorem, $APCD$ has a incircle . Similarly $BPDC$ has a incircle. Denote by $\omega_1 ,\omega_2$ ,the incircles of $\Delta APD$ and $\Delta BPD$ . Denote by $\Omega_1 ,\Omega_2$ ,the incircles of Quadrilaterals $APCD$ and $BPDC$ . Define $U \equiv DA \cap BC$ . Denote by $\Omega$ the $U$-excircle of $\Delta UAB$. Let $J$ denote the center of $\Omega$ . We show that $F,E,I,J$ lie on the same line . Denote by $f(\alpha, \beta)$ and $g(\alpha, \beta)$ the exsimilicenter and insimilicenter of circles $\alpha$ and $\beta$ respectively . Now we begin the Monge assault . First Monge's on the triples $\{\Omega,\Omega_1,\omega \}$ implies that $\underbrace {f(\Omega_1,\Omega)}_{A} - \underbrace {f(\Omega_1,\omega)}_{C} -{f(\Omega,\omega)}$ . Similarly $B-D-f(\Omega,\omega)$ . Hence $f(\Omega,\omega)= AC \cap BD \equiv E$ $\blacksquare$ . Next Monge on $\{\Omega,\omega_1,\omega \}$ implies $\underbrace {f(\Omega,\omega_1)}_{A} - \underbrace {g(\omega_1,\omega)}_{K} -{g(\Omega,\omega)}$ . Similarly $B-L-g(\Omega,\omega)$. Hence $g(\Omega,\omega)= AK \cap BL \equiv F$ $\blacksquare$ . Hence $I,J,F,E$ are collinear as desired

Solved with nukelauncher. Let $\omega_A$, $\omega_B$ denote the incircles of $\triangle APD$, $\triangle BPC$. Observe from $\triangle CPD$ and $\triangle APD$ that $$\frac{PC+PD-CD}2=PK=\frac{PA+PD-CD}2\implies PC+AD=PA+CD,$$ so by Pitot's theorem, quadrilateral $PADC$ has an incircle $\Gamma_A$. Analogously, quadrilateral $PBCD$ has an incircle $\Gamma_B$. [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=lightblue; pen sec=red; pen tri=purple; pen qua=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair P,C,D,I,K,L,IA,IB,A,B,EE,F,Q,O,I1,I2; P=dir(120); C=dir(205); D=dir(335); I=incenter(P,C,D); K=foot(I,P,D); L=foot(I,P,C); IA=K+0.8*(K-I); IB=extension(L,I,P,P+(C-P)*dir(90)*(IA-P)/(I-P)); A=extension(P,reflect(P,IA)*K,D,reflect(D,IA)*K); B=extension(P,reflect(P,IB)*L,C,reflect(C,IB)*L); EE=extension(A,C,B,D); F=extension(A,K,B,L); Q=extension(A,D,B,C); O=incenter(Q,A,B); I1=extension(A,IA,C,I); I2=extension(B,IB,D,I); draw(B--F--A,tri); draw(B--D,sec); draw(A--C,sec); draw(C--foot(O,B,C),pri); filldraw(incircle(Q,A,B),qfil,qua+Dotted); filldraw(circle(I1,abs(I1-foot(I1,C,D))),sfil,sec+dashed); filldraw(circle(I2,abs(I2-foot(I2,C,D))),sfil,sec+dashed); filldraw(incircle(P,A,D),tfil,tri); filldraw(incircle(P,C,B),tfil,tri); filldraw(incircle(P,C,D),qfil,qua); filldraw(A--B--C--D--cycle,fil,pri); draw(C--P--D,pri); draw( (EE+4*(EE-O))--(O+3*(O-EE)),blue+dashed); dot("$P$",P,N); dot("$C$",C,SW); dot("$D$",D,SE); dot("$A$",A,NE); dot("$B$",B,W); dot("$K$",K,dir(15)); dot("$L$",L,dir(120)); dot("$E$",EE,S); dot("$F$",F,dir(280)); dot("$I$",I,dir(260)); dot("$O$",O,SE); label("$\omega$",I+inradius(P,C,D)*SE,E,qua); label("$\Gamma$",O+inradius(Q,A,B)*SE,SE,qua); label("$\omega_A$",IA+inradius(A,D,P)*dir(300),dir(330),tri); label("$\omega_B$",IB+inradius(B,C,P)*dir(150),dir(150),tri); label("$\Gamma_A$",I1+abs(I1-foot(I1,C,D))*dir(-30),dir(120),sec); label("$\Gamma_B$",I2+abs(I2-foot(I2,C,D))*dir(150),W,sec); [/asy][/asy] If $I_A$, $I_B$ denote the centers of $\omega_A$, $\omega_B$, then let $O=\overline{AI_A}\cap\overline{BI_B}$. Then there is a circle $\Gamma$ centered at $O$ tangent to $\overline{AB}$, $\overline{BC}$, $\overline{AD}$. By Monge's theorem (with three exsimilicenters) on $\omega$, $\Gamma$, $\Gamma_A$, the exsimilicenter of $\omega$ and $\Gamma$ lies on $\overline{AC}$; analogously, it lies on $\overline{BD}$, so it is $E$. By Monge's theorem (with one exsimilicenter and two insimilicenters) on $\omega$, $\Gamma$, $\omega_A$, the insimilicenter of $\omega$ and $\Gamma$ lies on $\overline{AK}$; analogously, it lies on $\overline{BL}$, so it is $F$. Then the points $E$ and $F$ lie on line $OI$, as required.

Probably same as above$,$ but I$'$m extremely glad I managed to solve yet another $\text{G8}.$ Lemma 1: Quadrilaterals $APCD$ and $BPDC$ have incircles$.$ Proof: We only prove this for $APCD,$ as the argument will be similar for $BPDC.$ Let $w_1$ touch $AB, AD$ at $U_1, U_2$ respectively and $w$ touch $CD$ at $V.$ It is evident that $$AU_1=AU_2, PU=PK=PL, CL=CV, DV=DU_2.$$ Now $AP+CD=AD+CP$ and we are done$.$ $\hspace{1cm}\square$ Let the incircles of $APCD, BPDC$ be $k_1, k_2$ respectively$.$ Now for the main problem$:$ Let $X$ be the exsimilicenter of $w, w_1,$ $Y$ be the exsimilicenter of $w, w_2$ and $Z$ be the exsimilicenter of $w_1, w_2.$ By definition$,$ $X\in IK$ (since $IK$ is the line through the centers of $w, w_1$) and similarly $Y\in IL.$ First apply $\text{Monge's Theorem}$ to $w_1, w$ and $k_1.$ We easily see that $A$ is the exsimilicenter of $w_1$ and $k_1,$ $C$ is the exsimilicenter of $w$ and $k_1$ and $X$ is by definition the exsimilicenter of $w_1$ and $w.$ Thus due to the mentioned theorem$,$ the point $X$ must lie on $AC.$ By the exact same reasoning this time for $w, w_2$ and $k_2,$ we see that $Y$ lies on $BD.$ Now notice that $K$ is the insimilicenter of $w_1, w,$ $L$ is the insimilicenter of $w_2, w$ and $Z$ is the exsimilicenter of $w_1, w_2,$ therefore by $\text{Monge's Theorem},$ it follows that $K, L$ and $Z$ are collinear$.$ Another application of $\text{Monge's Theorem}$ to circles $w, w_1$ and $w_2$ yields that $X, Y$ and $Z$ are collinear$.$ Finally consider triangles $AKX$ and $BLY.$ We have that $$AB\cap KL\cap XY=Z.$$ Therefore $\text{Desargues's Theorem}$ tells us that $$F=AK\cap BL, I=KX\cap LY, E=AX\cap BY$$ are collinear$,$ as desired$.$ $\hspace{1cm}\blacksquare$

This problem is insane! Let $I_1$ be the incenter of $APD$, $I_2$ be the incenter of $BPC$ and $I$ be the incenter of $CPD$. Now, observe that since the incircles of $APD,PDC$ are tangent at point $K$, we have that $\frac{PC+PD-CD}{2}=PK= \frac{AP+PD-AD}{2} \implies AD+PC=AP+CD$, so by Pitot's Theorem, $APCD$ has an incircle $\omega_B$. Similarly, $BPDC$ has an incircle $\omega_A$. Let $K_B,K_A$ be the excimilienters of $\omega,\omega_1$ and $\omega,\omega_2$, respectively, where $\omega_1$ is the incircle of $APD$, $\omega_2$ is the incircle of $BPC$ and $\omega$ is the incenter of $CPD$. By Monge's Theorem on $\omega_B, \omega_1,\omega$, we have that $A,C,K_B$ are collinear. Similarly $B,D,K_A$ are collinear. Notice that since $K$ is the inscimilicenter of $\omega_1,\omega$, we have that $I,I_1,K,K_B$ are collinear, and moreover, $(I,I_1;K,K_B)=-1$. Similarly, $(I,I_2,L,K_A)=-1 \implies (I,I_2,L,K_A)=(I,I_1;K,K_B)$. Thus, by the so forgotten Prism Lemma, we have that $K_BK_A,I_1I_2,KL$ are concurrent. However, since $I_1I_2 \cap K_AK_B$ lies on $AB$ (due to Monge's Theorem on $\omega_1,\omega_2,\omega$, the concurrency point is actually the excimilicenter of $\omega_1,\omega_2$), so triangles $AK_BK,BK_AL$ are in perspective, so by Desargues' Theorem, we have that $AK_B \cap BK_A=E, AK \cap BL= F, K_BK \cap K_AL=I$ are collinear, so we are done! $\blacksquare$

Denote by $\Gamma$ circle, inscribed in both angles $ABC,BAD,$ by $\omega_1,\omega_2$ incircles of $APD,BPC.$ Note that $PBCD$ is tangential since $$|CD|+|CP|-|PD|=2|CL|=|BC|+|CP|-|PB|.$$ Then by the three homothety centers theorem center of external homothety $\omega\mapsto \omega_2$ lies on $BD,$ and analogously for center of external homothety $\omega\mapsto \omega_1.$ Applying three homothety centers theorem again, we can conclude that points $E,F$ are respectively centers of external and internal homothety $\omega\mapsto \Gamma,$ finally implying $I\in EF$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; 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 = -8.661033188860394, xmax = 9.796622553109685, ymin = -3.057994928096622, ymax = 6.113152768694817; /* image dimensions */ pen qqzzff = rgb(0,0.6,1); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1); pen qqqqcc = rgb(0,0,0.8); pen qqzzcc = rgb(0,0.6,0.8); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); draw((-1.5830860824502149,5.739481315434459)--(-2.692128000908574,0.507679826139392)--(-0.03245869884812526,-1.8102802942447922)--(2.9860163659991943,4.193898946519175)--cycle, linewidth(0.4) + xdxdff); /* draw figures *//* special point *//* special point *//* special point */ draw(circle((-0.011215221393844288,1.707172869407228), 1.560930244173962), linewidth(0.4) + qqzzff); /* special point */ draw(circle((-0.7515164232739878,4.2798836439913295), 1.116173900077119), linewidth(0.4)); /* special point *//* special point */ draw(circle((-1.9888002577400228,0.7427281455339743), 0.6392961535831577), linewidth(0.4)); /* special point *//* special point */ draw((-1.5830860824502149,5.739481315434459)--(-2.692128000908574,0.507679826139392), linewidth(0.4) + xdxdff); draw((-2.692128000908574,0.507679826139392)--(-0.03245869884812526,-1.8102802942447922), linewidth(0.4) + xdxdff); draw((-0.03245869884812526,-1.8102802942447922)--(2.9860163659991943,4.193898946519175), linewidth(0.4) + xdxdff); draw((2.9860163659991943,4.193898946519175)--(-1.5830860824502149,5.739481315434459), linewidth(0.4) + xdxdff); draw((-2.692128000908574,0.507679826139392)--(2.9860163659991943,4.193898946519175), linewidth(0.4)); draw((-1.5830860824502149,5.739481315434459)--(-0.03245869884812547,-1.8102802942447926), linewidth(0.4)); /* special point *//* special point */ draw(circle((-0.6238467778454819,1.1988881998148306), 1.879983042437203), linewidth(0.4) + qqqqcc); draw(circle((-0.0036057933673060358,2.9671269773510742), 2.120056515899163), linewidth(0.4) + qqzzcc); /* special point */ draw(circle((0.7434688451636409,1.6558376695206374), 3.122817022774604), linewidth(0.4) + yqqqyq); draw((-0.7654371025350631,1.758476630468249)--(0.7434688451636409,1.6558376695206374), linewidth(0.4) + dotted); draw((-1.5830860824502149,5.739481315434459)--(-0.44286019056493237,3.207234817383469), linewidth(0.4)); draw((-2.692128000908574,0.507679826139392)--(-1.4141946133352048,1.0229564906462136), linewidth(0.4)); draw((2.9860163659991943,4.193898946519175)--(-2.228813147027906,2.6933243019720856), linewidth(0.4)); draw((-2.228813147027906,2.6933243019720856)--(-0.03245869884812547,-1.8102802942447926), linewidth(0.4)); /* dots and labels */ label("$D$", (2.9471018676129757,4.137606802562069), NE * labelscalefactor); label("$P$", (-2.301793984009765,2.6235022299785773), SW * labelscalefactor); label("$I$", (-0.03784714690874773,1.6429392686864108), NE * labelscalefactor); label("$\omega$", (0.30823389825319125,3.070523579979418), NE * labelscalefactor,qqzzff); label("$K$", (-0.5137085840064138,3.142623797721489), NE * labelscalefactor); label("$\omega_1$", (-1.2635508485239482,5.2623701993383785), NE * labelscalefactor); label("$A$", (-1.6528920243311296,5.6805514622423905), NE * labelscalefactor); label("$L$", (-1.451011414653332,0.979617265459357), NE * labelscalefactor); label("$\omega_2$", (-2.2441138098161084,1.3401183541697124), NE * labelscalefactor); label("$C$", (-0.12436740819923246,-1.8899714006750712), SE * labelscalefactor); label("$B$", (-2.7632353775590173,0.431655610619617), SW * labelscalefactor); label("$E$", (-0.8309495420715245,1.6717793557832392), NE * labelscalefactor); label("$F$", (0.19287354986587826,1.6285192251379965), NE * labelscalefactor); label("$\Omega_2$", (-1.8980327646541697,2.3927815332039497), NE * labelscalefactor,qqqqcc); label("$\Omega_1$", (-0.758849324329454,4.757668675143881), NE * labelscalefactor,qqzzcc); label("$O$", (0.6543149434151302,1.5996791380411681), NE * labelscalefactor); label("$\gamma$", (0.19287354986587826,4.555788065466082), NW * labelscalefactor,yqqqyq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $\omega,\omega_1,$ and $\omega_2$ denote the incircles of $\triangle CPD,\triangle APD,$ and $\triangle BPC,$ respectively. Also, let $\gamma$ be the circle tangent to rays $\overline{AB},\overline{BC},$ and $\overline{AD}.$ Notice $AP+CD=AD+CP$ by equal tangents so $APCD$ has an incircle by Pitot, $\Omega_1.$ Similarly, let $\Omega_2$ be the incircle of $BPDC.$ Moreover, denote the exsimilicenter and insimilicenter of $\Omega$ and $\Gamma$ by $\ell_e(\Omega,\Gamma)$ and $\ell_i(\Omega,\Gamma),$ respectively. By Monge on $\gamma,\omega,$ and $\Omega_1,$ we see that $\ell_e(\gamma,\omega)$ lies on $\overline{AC}.$ By Monge on $\gamma,\omega,$ and $\Omega_2,$ we see that $\ell_e(\gamma,\omega)$ lies on $\overline{BD}.$ Hence, $E=\ell_e(\gamma,\omega).$ By Monge on $\gamma,\omega,$ and $\omega_1,$ we see that $\ell_i(\gamma,\omega)$ lies on $\overline{AK}.$ By Monge on $\gamma,\omega,$ and $\omega_2$ we see that $\ell_i(\gamma,\omega)$ lies on $\overline{BL}.$ Hence, $F=\ell_i(\gamma,\omega).$ Therefore, $E$ and $F$ both lie on $\overline{IO}$ where $O$ is the center of $\gamma.$ $\square$

It's not that hard to notice that $ADCP$ and $DPBC$ are circumscriptible. Call their incircles $\omega_1$ and $\omega_2$, respectively. Denote $O$ the centre of $\omega$. Also, let the incircles of $(ADP)$ and $(BPC)$ be $\omega_a$ and $\omega_b$. Finally, consider $T_1$ and $T_2$ the exsimilicenters of $\omega_a$ and $\omega$, and $\omega_b$ and $\omega$, respectively. Apply Monge for $\omega, \omega_a, \omega_1$ and get that $T_1$ lies on both $AC$ and $OK$. Apply Monge for $\omega, \omega_b, \omega_2$ and get that $T_2$ lies on both $BD$ and $OL$. Apply Monge one last time for $\omega, \omega_a, \omega_b$ and get that $T_1T_2, AB$ and $KL$ are concurent. Last but not least, use Desargue's theorem for triangles $AT_1K$ and $BT_2L$, and, together with our previous observations, the conclusion follows immediately.

It is clear that $APCD$ and $BPDC$ are both tangential. Indeed, $AP-AD=PK-DK=CP-CD$ and similarly for $BPDC$. Let the incircle of $APCD$ be $\Omega_A$ and incircle of $BPDC$ be $\Omega_B$. Let $\omega_A$ be the incenter and incircle of $\triangle APD$ and $\omega_B$ be the incenter and incircle of $\triangle BPC$. Let $\Omega$ be the circle tangent to segment $AB$, ray $BC$, and ray $AD$. Let $F'$ and $E'$ be the insimilicenters and exsimilicenters of $\omega$ and $\Omega$, respectively. Using Monge on $\omega$, $\Omega$, $\omega_A$ gives $B,L,F'$ collinear $\omega$, $\Omega$, $\omega_B$ gives $C,K,F'$ collinear $\omega$, $\Omega$, $\Omega_A$ gives $A,C,E'$ collinear $\omega$, $\Omega$, $\Omega_B$ gives $B,D,E'$ collinear So $E=E'$, $F=F'$. We are done.

im not very smart (it was only around 30 minutes though)---Monge + Menelaus (intuitively though, this feels like it should work, which is nice. incircles and collinearity and no other weird conditions basically means you cant mess up) Let's use Monge to obtain some collinearities. Define $CI\cap AK=U$ and $DI\cap BL=V$. By Monge it's not hard to prove $RUV$ collinear. This is all we need. Consider applying Menelaus on: $$\triangle AUC, \overline{EIF}$$ $$\triangle RAU, \overline{BVF}$$ $$\triangle RAC, \overline{EBD}$$ $$\triangle RUC, \overline{DIV}$$ where the first one is what we want to prove, and the rest are known collinearities. These give equations $$\frac{EA}{EC}\cdot \frac{IC}{IU}\cdot \frac{FU}{FA}=1$$ $$\frac{BR}{BA}\cdot \frac{VU}{VR}\cdot \frac{FA}{FU}=1$$ $$\frac{EC}{EA}\cdot \frac{BA}{BR}\cdot \frac{DR}{DC}=1$$ $$\frac{DC}{DR}\cdot \frac{IU}{IC}\cdot \frac{VR}{VU}=1$$ Since these multiply to $1$ and three of them hold, so the fourth must also hold and $EIF$ is collinear. Or just $\triangle AUC$ and $\triangle BVD$ perspective, oops.

Monge spam!!! This is a truly remarkable problem. I enjoyed this problem a lot! Let $\omega_A, \omega_B$ be the incircles of $ADP$ and $BCP$, respectively, and let $I_A, I_B$ be the centers of $\omega_A$ and $\omega_B$, respectively and denote $I$ by the center of $\omega$. By Monge-D-Alembert's theorem, we see that $AB, I_AI_B, KL$ are concurrent, so applying Desargues's theorem on triangles $BI_BL$ and $AI_AL$, we get that $I, F, BI_B \cap AI_A$ are collinear. Let $J = BI_B \cap AI_A$. Then $I, F, J$ are collinear. Thus it suffices to prove that the points $E, I, J$ are collinear. Let $\Omega$ be the circle tangents $BC, AB, AD$. Then it's clear that $J$ is the center of $\Omega$. Now consider the following claim: Claim: $AC \cap II_A$ is the exsimilicenter of $\omega$ and $\omega_A$. Proof. Note that $PC + PD - CD = 2PK = PD + AP - AD$, or equivalently, $AD + PC = CD + AP$. Thus quadrilateral $APCD$ is circumscribed about a circle $\gamma$. Applying Monge's theorem on circles $\omega, \omega_A, \gamma$ yields exsimilicenter of $\omega, \omega_A$ lies on line $AC$. Therefore $AC \cap II_A$ is the exsimilicenter of $\omega$ and $\omega_A$. $\blacksquare$ Now applying Monge theorem on circles $\Omega, \omega_A, \omega$, we see that the points $AC$ passes through the exsimilicenter of $\omega$ and $\Omega$. Similarly $BD$ passes through the exsimilicenter of $\omega$ and $\Omega$. Hence $E$ is the exsimilicenter of $\omega$ and $\Omega$, which means that $I, J, E$ are collinear. Thus we're done. $\blacksquare$

Straightforward. Claim 1: $APCD,BCPD$ are inscribed quadrilateral.

Let $\omega$ be the incircle of $GAB$ where $G = AB \cap CD$ Let $\Omega$ be the incircle of $PCD$ Let $\omega_1$ denote the incircle of $BPCD$ Let $\omega_2$ denote the incircle of $APCD$ Let $\Gamma_1$ denote the incircle of $BPC$ Let $\Gamma_2$ denote the incircle of $APD$ Claim 2: $F$ is the insimilicenter of $\omega$ and $\Omega$

Claim 3: $E$ is the exsimilscenter of $\omega$ and $\Omega$

So we're done.

From a simple length chase, one deduces that $BPDC$ and $APCD$ have incircles. Let the incentres be $I_a, I_b$ respectively. Then note that if we introduce $X \equiv AC \cap BD$, it holds that $I_a, I_b,$ and $X$ are collinear. Moreover, if we let $I_1, I_2$ denote the incentres of $\triangle APD$ and $\triangle BPC$, we have $I_a = BI_2 \cap DI$ and $I_b = AI_1 \cap CI$. Now we note that $\triangle AI_bC$ and $\triangle BI_aD$ are perspective, (due to perspector $Y$) and thus they have a perspectrix, by Desargue's. This implies that the following points are collinear $$AI_1 \cap BI_2 \overset{\text{def}}{=} R, CI_1 \cap DI_2 \equiv I, AC \cap BD \equiv E$$ Now, let $\mathcal P_{I}$ denote the pencil of lines through $I$. Consider the involution $\tau : \mathcal P_{I} \mapsto \mathcal P_{I}$ which swaps the following pairs $$(IA, II_2) \text{ and } (IB, II_1)$$ From DDIT, this also swaps $(IR, IX)$, where $X = AB \cap I_1I_2$. Now note that from Monge, on incircles of $\triangle PAD, \triangle PDC, \triangle PCB$ where we take two insimilicentres and one exsimilicenter, we obtain that $K, L,$ and $X$ are collinear. Since $II_2 \equiv IL$ and $II_1 \equiv IK$, from DDIT again, we see that the involution swaps $(IF, IX)$. As a result, we obtain $IF \equiv IR$, or that $R, I, F$ collinear. Since $R, I, E$ collinear from before, we obtain that $E, I, F$ are collinear; which is precisely what we wanted to show.

From an easy length chase, we see that $BP + CD = BC + PD$, so $BPCD$ has an inscribed circle which will be denoted as $\omega_1$. Similarly, $APCD$ has an inscribed circle, $\omega_2$. Then let $AD \cap BC = X$. Let the inscribed circles of $\triangle XAB$, $\triangle PCD$, $\triangle PBC$, $\triangle PDA$ be $\Omega$, $\gamma$, $\gamma_1$, and $\gamma_2$ respectively. Apply Monge D'Alembert on $(\gamma_1, \Omega, \gamma)$ to obtain that $BL$ passes through the insimilicenter of $\gamma$ and $\Omega$, and similarly for $AK$. So $AK \cap BL$ is the insimilicenter. Now apply Monge's on $(\gamma, \Omega, \omega_1)$ to get that the exsimilicenter of $\Omega$ and $\omega$ lies on $BD$ and similarly $AC$. Thus $E$ is the desired exsimilicenter. Now apply Monge D'Alembert on $(\gamma, \gamma, \Omega)$ to get that $I \in EF$ as desired.

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.53303145358391cm); 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 = -11.734353072407469, xmax = 10.798678381176442, ymin = -7.840194452658065, ymax = 5.195774054650439; /* image dimensions */ pen qqqqcc = rgb(0.,0.,0.8); pen wwqqcc = rgb(0.4,0.,0.8); pen zzccff = rgb(0.6,0.8,1.); pen bcduew = rgb(0.7372549019607844,0.8313725490196079,0.9019607843137255); pen qqzzcc = rgb(0.,0.6,0.8); pen cczzff = rgb(0.8,0.6,1.); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen ttqqqq = rgb(0.2,0.,0.); /* draw figures */ draw((-3.0363103906755877,-3.4034824617611217)--(3.5792462648615238,-7.191277425544798), linewidth(1.) + qqqqcc); draw((-3.0363103906755877,-3.4034824617611217)--(3.6345724438368903,3.50373998588049), linewidth(1.) + qqqqcc); draw((3.6345724438368903,3.50373998588049)--(3.5792462648615238,-7.191277425544798), linewidth(1.) + wwqqcc); draw(circle((1.0602552330014576,-2.820326714276643), 2.5415683019567763), linewidth(1.) + zzccff); draw(circle((-2.1868828578625514,0.3157063172109016), 1.9726989648010216), linewidth(1.) + bcduew); draw(circle((-0.8243270164891944,-6.111835563813467), 1.25127769566486), linewidth(1.) + bcduew); draw((-1.7170106805997845,-7.399105701272271)--(-4.757450125732407,1.8091510826242967), linewidth(1.) + wwqqcc); draw((-4.757450125732407,1.8091510826242967)--(3.6345724438368903,3.50373998588049), linewidth(1.) + wwqqcc); draw((-1.7170106805997845,-7.399105701272271)--(3.5792462648615238,-7.191277425544798), linewidth(1.) + wwqqcc); draw((-1.7170106805997845,-7.399105701272271)--(3.6345724438368903,3.50373998588049), linewidth(1.) + qqzzcc); draw((-4.757450125732407,1.8091510826242967)--(3.5792462648615238,-7.191277425544798), linewidth(1.) + qqzzcc); draw((1.3578459761495199,-2.5806750089835853)--(0.18414028001669538,-3.5258676202015207), linewidth(1.) + dotted); draw((-1.7170106805997845,-7.399105701272271)--(1.3578459761495199,-2.5806750089835853), linewidth(1.) + cczzff); draw((-4.757450125732407,1.8091510826242967)--(1.3578459761495199,-2.5806750089835853), linewidth(1.) + cczzff); draw(circle((-0.022306754756019308,-0.9418662392163756), 3.6338331009132188), linewidth(1.) + dotted + eqeqeq); draw(circle((0.49460229307766634,-4.2099095155574275), 3.100025346787125), linewidth(1.) + dotted + eqeqeq); draw(circle((1.9615959046924973,-2.094471377730541), 5.156315228293298), linewidth(1.) + linetype("2 2")); draw((1.3578459761495199,-2.5806750089835853)--(1.9615959046924973,-2.094471377730541), linewidth(1.) + dotted + ttqqqq); /* dots and labels */ dot((-3.0363103906755877,-3.4034824617611217),dotstyle); label("$P$", (-3.496900580059136,-3.661486756296904), NE * labelscalefactor); dot((3.5792462648615238,-7.191277425544798),dotstyle); label("$C$", (3.7808487092971585,-7.240379950788042), NE * labelscalefactor); dot((3.6345724438368903,3.50373998588049),dotstyle); label("$D$", (3.7208672591101557,3.696237799975381), NE * labelscalefactor); dot((1.0602552330014576,-2.820326714276643),linewidth(4.pt) + dotstyle); label("$I$", (0.8017700166760379,-2.7017835533048666), NE * labelscalefactor); dot((-0.7679092854329346,-1.054715226137477),linewidth(4.pt) + dotstyle); label("$T_1$", (-0.8977044052890308,-0.7623833305917913), NE * labelscalefactor); dot((-0.20259448742994288,-5.025951346238707),linewidth(4.pt) + dotstyle); label("$T_2$", (-0.257902269961005,-5.600886979009979), NE * labelscalefactor); dot((-4.757450125732407,1.8091510826242967),linewidth(4.pt) + dotstyle); label("$A$", (-4.676535767070184,1.9767695612813148), NE * labelscalefactor); dot((-1.7170106805997845,-7.399105701272271),linewidth(4.pt) + dotstyle); label("$B$", (-2.0773395923000786,-7.320355217704045), NE * labelscalefactor); dot((0.18414028001669538,-3.5258676202015207),linewidth(4.pt) + dotstyle); label("$E$", (0.06199879770300793,-3.281604238445889), NE * labelscalefactor); dot((1.3578459761495199,-2.5806750089835853),linewidth(4.pt) + dotstyle); label("$F$", (1.3216092516300588,-2.3618886689118535), NE * labelscalefactor); dot((1.9615959046924973,-2.094471377730541),linewidth(4.pt) + dotstyle); label("$J$", (2.041386653874088,-1.942018517602837), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] 2007 G8 Here is a sketch of the problem. Firstly we show that $APCD$ and $BPCD$ are tangential by pitot theorem. After that consider a circle $\Omega$ which is tangent to $AB, AD$ and $CB$ and let $O$ denote its center, let $J$ denote the insimilicenter of $\omega$ and $\Omega$. Let $T_1$ and $T_2$ denote the tangency points of $\omega$ to $PC$ and $PD$. Then by Monge's theorem, we have that $J$, $T_1$ and $B$ are collinear. Similarly $J, T_2$ and $A$ are collinear. hence $J=F$. Now we apply Monge's theorem again, we show that $E$ is the exsimilicenter of $\Omega$ and $\omega$. hence $E-I-F$ are collinear.

Let the incircles of $\triangle APD, \triangle BPC$ be $\omega_1, \omega_4$ respectively, now it's known that by a corolary of Pitot's theorem we get that there exists incircles $\omega_2, \omega_3$ of $APCD, BPDC$ respectively. Because of: $$AD+DP-AP=2DK=CD+DP-PC \implies AD+PC=CD+AP$$ And similarily for the other quadrilateral, so now by Monge d'Alembert 2 times+ notising the angle bisectors we get that $CI \cap AK=I_2, DI \cap BL=I_3$ are the insimillicenters of $(\omega_2, \omega)$ and $(\omega_3, \omega)$ respectively. And by Monge'd Alembert one more time we get that $AB,CD,I_2I_3$ are concurrent at point $G$, now from Desargues perspective theorem on $\triangle AI_2C, \triangle BI_3D$ we get that $E,I,F$ are colinear as desired thus we are done .