such that:
such that 
,
is the Gergon point and the incircle 
touch 
,
,
,
respectively.
,
,
be 
, 
, 
respectively.
and then we get the circle 
and then the circle 
and then the circle 
be non-negative real numbers such that 
Prove that
Y by Rounak_iitr, GeoKing, jrpartty, ehuseyinyigit
Line 
intersects sides 
and 
of cyclic quadrilateral 
in its interior points 
and 
, respectively, and intersects ray 
beyond point 
at 
, and ray 
beyond point 
at 
. Circumcircles of the triangles 
and 
intersect at 
, while circumcircles of the triangles 
and 
intersect at 
. Let lines 
and 
meet at point 
, lines 
and 
meet at point 
and lines and meet at point 
. Prove that point lies on line 
.
intersects sides 
and 
of cyclic quadrilateral 
in its interior points 
and 
, respectively, and intersects ray 
beyond point 
at 
, and ray 
beyond point 
at 
. Circumcircles of the triangles 
and 
intersect at 
, while circumcircles of the triangles 
and 
intersect at 
. Let lines 
and 
meet at point 
, lines 
and 
meet at point 
and lines and meet at point 
. Prove that point lies on line 
.![[asy]
size(10cm);
import geometry;
defaultpen(fontsize(10pt));
pair A,B,C,D,R,S,M,P,N,Q,T,U,X,Y;
A=dir(135);
D=dir(85);
C=dir(-20);
B=dir(200);
R=(2*A+B)/3;
S=(5*C+2*D)/7;
P=extension(A,D,R,S);
Q=extension(B,C,R,S);
X=extension(A,B,C,D);
Y=extension(A,D,C,B);
M=2*foot(A,circumcenter(P,A,R),circumcenter(X,A,D))-A;
N=2*foot(C,circumcenter(Q,C,S),circumcenter(X,C,B))-C;
T=foot((0,0),X,Y);
U=extension(R,S,X,Y);
draw(unitcircle);
draw(circle(X,A,D));
draw(circle(X,B,C));
draw(circle(X,R,S));
draw(circle(P,A,R));
draw(circle(S,C,Q));
draw(circle(T,U,P),blue+dotted);
draw(circle(T,U,Q),blue+dotted);
draw(circle(M,N,P),blue+dotted);
draw(X--extension(M,P,X,Y),green+dotted);
draw(P--extension(M,P,X,Y),green+dotted);
draw(Q--extension(M,P,X,Y),green+dotted);
draw(U--Q);
draw(X--B--C--cycle);
draw(D--Y--B);
draw(X--Y);
draw(C--Q);
dot("$A$",A,SE);
dot("$D$",B,SW);
dot("$C$",C,dir(270));
dot("$B$",D,NE);
dot("$S$",R,dir(270));
dot("$R$",S,dir(270));
dot("$P$",P,SW);
dot("$Q$",Q,ESE);
dot("$L$",X,dir(90));
dot("$K$",Y,dir(270));
dot("$T$",T,dir(T));
dot("$U$",U,dir(T));
dot("$M$",M,NE);
dot("$N$",N,W);
clip(currentpicture, (-4,-1.1)--(3,-1.1)--(3,3)--(-4,3)--cycle);
[/asy]](http://latex.artofproblemsolving.com/0/4/9/049144054b7014af03d797403f018d0658b7082a.png)
and 
.
,
(radax), and let 
.
,
,
exist, after which the problem finishes by radax.
,
are cyclic. Similarly 
are cyclic.
,
are cyclic, as desired.
and move 
,
are fixed, 
,
,
.
to a fixed line and sends 
to a variable line that stays parallel to a fixed line, so the image of 
.
we have 
,
we have 
(assume these cases are distinct by continuity), so in fact 
.
we have 
,
are always collinear lines 
coincide. Thus 
and we want to show it lies on the fixed line 
,
are concurrent, all three lines 
share the point 
.
and 
.
at 
;
because 
,
.
;
.
this implies 
are collinear, i.e. 
concurrent as desired.
is the Miquel point of 
and 
.
be 
.
be the Miquel point of 
is cyclic.
both lie on 
is cyclic
.
and 
are cyclic.
.
.
is cyclic.
as 
.
is cyclic
.
.
.
and 
are cyclic
,
the result follows.
to a rectangle. Then show 
,
lie on the line at infinity.
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
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 = -8.5175300171852, xmax = 11.601175595567593, ymin = -8.956055083733087, ymax = 7.410766922176333; /* image dimensions */
pen ccqqqq = rgb(0.8,0,0);
/* draw figures */
draw(circle((4.254919711426577,-5.20298347684431), 3.7435131156477155), linewidth(0.4));
draw((xmin, 1.54261979760901*xmin-5.28438719926584)--(xmax, 1.54261979760901*xmax-5.28438719926584), linewidth(0.4)); /* line */
draw((xmin, -12.941176470588207*xmin + 89.11235294117625)--(xmax, -12.941176470588207*xmax + 89.11235294117625), linewidth(0.4)); /* line */
draw((xmin, -0.10546875000000003*xmin-2.021171875)--(xmax, -0.10546875000000003*xmax-2.021171875), linewidth(0.4)); /* line */
draw((xmin, -0.4144990785383048*xmin-4.086126855675013)--(xmax, -0.4144990785383048*xmax-4.086126855675013) , linewidth(0.4)); /* line */
draw((xmin, -0.5939595367280669*xmin-2.1697333752991956)--(xmax, -0.5939595367280669*xmax-2.1697333752991956), linewidth(0.4)); /* line */
draw(circle((0.8426595582669799,-2.064783447542708), 1.1492779427133082), linewidth(0.4));
draw(circle((3 .1232469441580526,-4.984028973908837), 4.551537011214573), linewidth(0.4));
draw(circle((9.531527920699311,-6.701979989357918), 2.1432526154016927), linewidth(0.4) + ccqqqq);
draw(circle((5.56367459938652,-6.623443014817272), 5.452620481752408), linewidth(0.4) + ccqqqq);
draw((xmin, 0.4610732271996606*xmin + 1.764487879216972)--(xmax, 0.4610732271996606*xmax + 1.764487879216972), linewidth(0.4)); /* line */
draw(circle((-3.0156517025322014,2.874013441845172), 5.567963936631288), linewidth(0.4) + ccqqqq);
draw(circle((-2.266863245489841,-0.9752515580494889), 4.428345185108061), linewidth(0.4) + ccqqqq);
draw(circle((1.32468546885748,-6.539539363532021), 9.559742688704302), linewidth(0.4) + ccqqqq);
draw(circle((0.25265982197443293,-4.214473444189836), 7.515905516100535), linewidth(0.4) + blue);
draw((xmin, 0.7812249313369485*xmin-1.7515075390816968)--(xmax, 0.7812249313369485*xmax-1.7515075390816968), linewidth(0.4)); /* line */
draw((xmin, 50.522105600837584*xmin-548.0197493633044)--(xmax, 50.522105600837584*xmax-548.0197493633044), linewidth(0.4)); /* line */draw((xmin, 3.7301884156842817*xmin-34.137848032365724)--(xmax, 3.7301884156842817*xmax-34.137848032365724), linewidth(0.4)); /* line */
draw((-6.682046355909883,-1.3164247984001296)--(7.3929280424441135,-6.560833490453235), linewidth(0.4));
draw((7.609109322964804,-5.75443658216716)--
(-6.682046355909883,-1.3164247984001296), linewidth(0.4));
draw((7.609109322964804,-5.75443658216716)--(2.5519080152048264,2.9411043433041417), linewidth(0.4));
draw((7.3929280424441135,-6.560833490453235)--(2.551908015204826 4,2.9411043433041417), linewidth(0.4));
draw(circle((6.733111604383132,-0.9128314619886586), 5.6864122824040315), linewidth(0.4) + blue);
draw(circle((-5.828642262913379,-23.797304463991193), 22.497072011369145), linewidth(0.4) + blue);
draw((7.609109322964804,-5.75443658216716)--(10.678639181615239,-8.512412956497565), linewidth(0.4));
draw((1.4577758821827405,-3.035593262933805)--(7.609109322964804,-5.75443658216716), linewidth(0.4));
draw((7.3929280424441135,-6.560833490453235)--(10.749381590183365,-4.938357520361102), linewidth(0.4));
/* dots and labels */
dot((1.98,-2.23),dotstyle); label("$A$", (2.0650556867798113,-2.002520878910299), NE * labelscalefactor);
dot((7.1,-2.77),dotstyle);
label("$B$", (7.1974261712918555,-2.5228533704276503), NE * labelscalefactor);
dot((7.44,-7.17),dotstyle);
label("$C$", (7.528546847711987,-7.631572378052557), NE * labelscalefactor);
dot((0.612257312 6215318,-4.339906947584976),dotstyle);
label("$D$", (0.7169215042121316,-4.107502321866857), NE * labelscalefactor);
dot((-0.30412344374858513,-1.9890963555421415),dotstyle);
label("$P$", (-0.2054860943868071,-1.742354633151623), NE * labelscalefactor );
dot((1.4577758821827405,-3.035593262933805),dotstyle);
label("$R$", (0.9534362730836544,-3.4925639228008962), NE * labelscalefactor);
dot((7.3929280424441135,-6.560833490453235),linewidth(4pt) + dotstyle);
label("$S$", (7.481243893937683,-6.378044103033482), NE * labelscalefactor);
dot((10.6786391816 15239,-8.512412956497565),linewidth(4pt) + dotstyle);
label("$Q$", (10.768799181251849,-8.317465207779975), NE * labelscalefactor);
dot((1.0467378714360498,-0.9337698173412851),linewidth(4pt) + dotstyle);
label("$M$", (1.1426480881808727,-0.7489926038912246), NE * labelscalefactor);
dot((10.749381590183365,-4.938357 520361102),linewidth(4pt) + dotstyle);
label("$N$", (10.839753611913306,-4.74609219781997), NE * labelscalefactor);
dot((6.51740319958198,4.769488005409629),linewidth(4pt) + dotstyle);
label("$K$", (6.606139249113048,4.9510133259124895), NE * labelscalefactor);
dot((-6.682046355909883,-1.316424798 4001296),linewidth(4pt) + dotstyle);
label("$L$", (-6.591384853917922,-1.1274162340856622), NE * labelscalefactor);
dot((0.8596528747804995,2.1608508044634824),linewidth(4pt) + dotstyle);
label("$T$", (0.9534362730836544,2.349350868325732), NE * labelscalefactor);
dot((2.5519080152048264,2.9411043433041417),linewidth(4pt) + dotstyle) ;
label("$E$", (2.656342608958618,3.1298496056017595), NE * labelscalefactor);
dot((7.609109322964804,-5.75443658216716),linewidth(4pt) + dotstyle);
label("$L'$", (7.694107185922054,-5.573893888870303), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/a/c/2/ac2ca54e8aaa4fb77dbbfe1f64482675312acb4d.png)
Let 
is cyclic
are cyclic
similarly other follows
is cyclic
is cyclic
is cyclic by miquel, now 
so 
is cyclic.
is cyclic
to meet 
at 
then by Reims theorem we have 
h
so done.
,
meet 
,
.
,
and similarly 
.
,
and 
are isogonal, and 
.
and 
are homothetic, the concurrency follows.
is the Miquel point of 
concur regardless of whether 
.![\[\measuredangle BMA = \measuredangle BMP + \measuredangle PMA = \measuredangle BRP + \measuredangle PSA\]](http://latex.artofproblemsolving.com/e/7/0/e7078003058ebe851fb2a736034390d0cfb64508.png)
![\[=\measuredangle LRS + \measuredangle RSL = \measuredangle RLS = \measuredangle BLA \]](http://latex.artofproblemsolving.com/7/d/9/7d920f8b2761a3f97245a3f025c7da8283b25b01.png)
So it follows that 
.
.
and 
.![\[\measuredangle LRN = \measuredangle CRN = \measuredangle DSN = \measuredangle LSN \]](http://latex.artofproblemsolving.com/9/6/9/969cf2afe459efbc557afbc7aa0f620018d367ec.png)
from which it is clear that 
.
is cyclic![\[ \measuredangle QNM = \measuredangle QNR + \measuredangle RNM = \measuredangle DCR+ \measuredangle PSM = \measuredangle SAP + \measuredangle PSM \]](http://latex.artofproblemsolving.com/b/0/f/b0f6305c30a0818cf3ec99629da3771babc9420c.png)
![\[ = \measuredangle SMP + \measuredangle PSM = \measuredangle SPM = \measuredangle QPM \]](http://latex.artofproblemsolving.com/2/1/7/2176bf05a6b2ab63a46182df378a9343c5a083aa.png)
is cyclic. ![\[\measuredangle NTM = \measuredangle NTL + \measuredangle LTM = \measuredangle NCL + \measuredangle LAM = \measuredangle NCR + \measuredangle LAM \]](http://latex.artofproblemsolving.com/f/4/7/f47a9029b067d77a0098dbc3ec6e9239173cddf3.png)
![\[ = \measuredangle NQR + \measuredangle SPM = \measuredangle XQP + \measuredangle QPX = \measuredangle QXP \]](http://latex.artofproblemsolving.com/6/4/3/643572949f6410f4482f6176d180e081b289829f.png)
which implies that indeed 
as claimed.
so we show 
.![\[\measuredangle NTX = \measuredangle NMX = \measuredangle XQP = \measuredangle XQR = \measuredangle NCR = \measuredangle NTL\]](http://latex.artofproblemsolving.com/8/2/c/82cde5221a739f98ca138088519f56aec1a2b0ad.png)
from which it is clear that points 
and 
.
and 
.![\[180-\angle FLS=180-\angle FLD=\angle DNF=180-\angle QND=180-\angle QSD=\angle LSQ\]](http://latex.artofproblemsolving.com/a/b/c/abcaeac59909296eb5f7f961520c0de1146047f8.png)
Thus, 
.![\[180-\angle ELS=180-\angle ELA=180-\angle EMA=\angle AMP=180-\angle PSA=\angle ASR=\angle LSR \]](http://latex.artofproblemsolving.com/5/2/0/520f88a5e8fc47c55d9131f673ab27084b19ff06.png)
So 
.
are collinear and 
.![\[\angle NMP=\angle NMS+\angle SMP=180-\angle SRN+\angle SAP=\angle NRQ+\angle DCB=\angle NCQ+\angle DCB=180-\angle RQN\]](http://latex.artofproblemsolving.com/b/7/8/b78c94c1821457ddc8de534880ccfbf1aa035c00.png)
This yields 
are cyclic. We get![\[\angle EFN=\angle EFQ=\angle PQF=\angle PQX=\angle XMN=\angle EMN\]](http://latex.artofproblemsolving.com/4/e/c/4ec69bd01b7a3c9d0c899d9bafc5ba4c5056334f.png)
Hence 
are cyclic. We have 
thus, 
,
is cyclic. Likewise, 
so 
is cyclic. Therefore, 
is cyclic. Now, we claim that 
is cyclic. To that end, we have
.
intersect 
for the second time at 
respectively. We will show that triangles 
and 
so 
.
and 
.
)
and 
such that 
and 
and 
and 
so 
cyclic and since 
cyclic by reim it is easy to get the first conclusion.
collinear in this order(other case similar) and let 
.
.
the angle 
and 
we want to show that 
but the latter is just 
and now angle chase and law of sines solves the problem (basicaly now we are working only with 
)
are concyclic because 
and 
.
are concyclic by angle chasing: ![\[\measuredangle NMP = \measuredangle NMS + \measuredangle SMP = \measuredangle NRQ + \measuredangle DAB = \measuredangle NCQ + \measuredangle QCB = \measuredangle NQP\]](http://latex.artofproblemsolving.com/2/b/8/2b87834dd4f97307fd611194eb1024929f9bbf0c.png)
Let 
intersect line 
are concyclic. Indeed, this is again by angle chasing: ![\[\measuredangle KTM = \measuredangle LTM = \measuredangle LRM = \measuredangle BRM = \measuredangle BPM = \measuredangle KPM\]](http://latex.artofproblemsolving.com/2/b/9/2b92602a3ee77d0290ac0acb2e3fa14707eae217.png)
Similarly, we get that 
are concyclic. Now applying the Radical Axis theorem on 
gives us the desired result.
meet 
meet 
are concyclic, implying 
are concyclic, so are 
.
are concyclic.
,
as the intersection of 
and 
;
are harmonic pencil.
we have![\[
\frac{\sin\angle ABD}{\sin\angle DBC}\cdot \frac{\sin\angle BCA}{\sin\angle ACD}\cdot \frac{\sin\angle CKG}{\sin\angle GKA} = 1
\]](http://latex.artofproblemsolving.com/b/8/3/b837ccc925d7458a26684b588e940be9c5f38d8d.png)
![\[
\frac{\sin\angle LKA}{\sin\angle CKL}
= \frac{\sin\angle GKA}{\sin\angle CKG}
= \frac{\sin\angle ABD}{\sin\angle DBC}\cdot \frac{\sin\angle BCA}{\sin\angle ACD}
=\frac{AD}{CD}\cdot \frac{AB}{AD}
=\frac{AB}{CD}
\]](http://latex.artofproblemsolving.com/7/0/5/70586c50a16567b37b07d1ee8e59691deb89fda3.png)
and 
which are concurrent. By spiral similarity, we have triangles 
,
,![\[
\frac{\sin\angle MPQ}{\sin\angle MPK} =
\frac{\sin\angle MPS}{\sin\angle MPA}
=\frac{MS}{MA}
=\frac{RS}{AB}
\]](http://latex.artofproblemsolving.com/e/f/c/efc8854471dd35b99391e4a065cd11c2fbb56de9.png)
where the middle equality is by considering the length of arcs subtended on circle 
while the last is due to the triangle similarity.![\[
\frac{\sin\angle NQK}{\sin\angle NQP} =
\frac{\sin\angle NQC}{\sin\angle NQR}
=\frac{NC}{NR}
=\frac{CD}{RS}
\]](http://latex.artofproblemsolving.com/5/5/c/55c3a883be0591b1f5da026908b33c8d492a5f4a.png)
Therefore, by Ceva's theorem, we have![\[
\frac{\sin\angle XKP}{\sin\angle QKX}
= \frac{\sin\angle MPK}{\sin\angle MPQ}\cdot \frac{\sin\angle NQP}{\sin\angle NQK}
=\frac{AB}{RS}\cdot \frac{RS}{CD}
=\frac{AB}{CD}
\]](http://latex.artofproblemsolving.com/c/5/d/c5d9aa411cbbb03b23eeed78ece0ab92851a0341.png)
so 
.
:
is in angle domain 
but not both. Thus 
.
so we need to prove that 
.
and 
,
and 
are cyclic.
is cyclic
is cyclic.
,
and 
.
.
,
is cyclic in particular. Thus, 
and 
,
be the Miquel point of 
and let 
be the Miquel point 
.
,
and 
we obtain that 
concur at 
.
and 
.
,
and similarly 
,
are collinear and 
.
,
is cyclic. Using 
,
is cyclic by Reim's theorem.
.
,
,
we obtain that 
concur at 
,![\[ \frac{\sin \angle KPX}{\sin \angle XPQ} = \frac{MA}{MS} = \frac{AB}{SR} \]](http://latex.artofproblemsolving.com/d/2/5/d2579a504eeaa65fedac7844044f97977b8678f7.png)
and similarly 
so by Trig Ceva on 
we have![\[ \frac{\sin \angle PKX}{\sin \angle XKQ} =\frac{\sin \angle KPX}{\sin \angle XPQ} \cdot \frac{\sin \angle XQP}{\sin \angle KQX} = \frac{AB}{SR} \cdot \frac{SR}{CD} = \frac{AB}{CD}. \]](http://latex.artofproblemsolving.com/e/9/8/e98fbfdd8e5c93f496247b60820a4a329de1ec66.png)
![\[ \frac{\sin \angle PKL}{\sin \angle LKQ} = \frac{BL}{LC} \cdot \frac{KC}{KB} = \frac{BL}{LC} \cdot \frac{AC}{BD} = \frac{LB}{BD} \cdot \frac{AC}{LC} = \frac{LA}{AC} \cdot \frac{AC}{LC} = \frac{LA}{LC} = \frac{AB}{CD} \]](http://latex.artofproblemsolving.com/7/2/4/724e5e508b2867f7bc0aa989ba5e95652d9dfd7b.png)
which solves the problem (ignoring sign issues).
intersect at 
.
are concurrent.
that's true coz 
applying trigonometric Ceva on 
and 
which clearly multiply 1. We are done
and 
.
and 
.![\[ \angle MUN = \angle MUR - \angle NUR = \angle MPR - \angle NQR = \angle MXN. \]](http://latex.artofproblemsolving.com/8/9/6/896d603a9554b6d15b8c0fca208b032746e924bd.png)
The same angle chase also gives 
,![[asy]
defaultpen(fontsize(10pt));
size(400);
pair A, B, C, D, P, Q, R, S, M, N, X, K, L, E, F, O, U, V, W;
A = dir(140);
B = dir(80);
C = dir(340);
D = dir(200);
R = 0.65 * A + 0.35 * D;
S = 0.2 * B + 0.8 * C;
P = extension(A, B, R, S);
Q = extension(C, D, R, S);
M = IP(circumcircle(P, A, S), circumcircle(P, B, R), 0);
N = IP(circumcircle(Q, C, R), circumcircle(Q, D, S), 0);
X = extension(M, P, N, Q);
K = extension(A, B, C, D);
L = extension(B, C, A, D);
E = extension(A, C, B, D);
F = foot(E, K, L);
O = (0,0);
U = IP(circumcircle(P, B, R), circumcircle(Q, C, R), 0);
V = IP(circumcircle(P, A, S), circumcircle(Q, D, S), 1);
W = extension(U, R, S, V);
draw(A--B--C--D--cycle, orange);
draw(circumcircle(A, B, C), red);
draw(A--P--Q--C, orange);
draw(M--X--Q, lightblue);
draw(B--L--A--K--D, orange);
draw(K--L, heavycyan);
draw(circumcircle(X, M, N), dashed+magenta);
draw(circumcircle(P, A, S), dotted+heavygreen);
draw(circumcircle(P, B, R), dotted+heavygreen);
draw(circumcircle(Q, C, R), dotted+heavygreen);
draw(circumcircle(Q, D, S), dotted+heavygreen);
draw(circumcircle(F, A, D), purple+dotted);
draw(circumcircle(F, B, C), purple+dotted);
clip((K-(0.2,1))--(Q+(0.2,-1))--(Q+(0.2,3))--(K+(-0.2,3))--cycle);
dot("$A$", A, dir(A));
dot("$B$", B, dir(70));
dot("$C$", C, dir(300));
dot("$D$", D, dir(245));
dot("$P$", P, dir(120));
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(25));
dot("$S$", S, dir(70));
dot("$M$", M, dir(20));
dot("$N$", N, dir(40));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$X$", X, dir(150));
dot("$F$", F, dir(100));
dot("$U$", U, dir(150));
dot("$V$", V, dir(220));
[/asy]](http://latex.artofproblemsolving.com/b/c/4/bc4e490b389927ea7ebca58f3ab0fb1011556986.png)
.
.![\[ \angle UBK = \angle UBP = \angle UMX = \angle UNX = 180^\circ - \angle UNQ = \angle UCK. \]](http://latex.artofproblemsolving.com/5/e/e/5ee666f3f5e1783b7c293fe47d747c44aa671a4c.png)
The claim for 
is Miquel point of 
.![\[ \angle UFV = \angle UFK + \angle VFK = \angle UCK + \angle VDK = \angle UNX + \angle VNX = \angle UNV, \]](http://latex.artofproblemsolving.com/b/b/c/bbc0d885672804f51e6abc9620168c7a9b5aa06b.png)
which proves the claim. 
,
,
.
