Anyone have an idea?

Iran 2nd Round 2021/3 wrote: Circle $\omega$ is inscribed in quadrilateral $ABCD$ and is tangent to segments $BC, AD$ at $E,F$ , respectively.$DE$ intersects $\omega$ for the second time at $X$. if the circumcircle of triangle $DFX$ is tangent to lines $AB$ and $CD$ , prove that quadrilateral $AFXC$ is cyclic. Nice problem! Let $\omega$ tangent to $AB$ and $CD$ at $L$ and $K$ respectively. Let $(DFX) = \Omega$. Furthermore, let $FX$ intersects $CD$ at $M_1$, and define $Z \in CD$ to be the point such that $AZ \parallel FX$. Claim 01. $EFDC$ is an isosceles trapezoid, with $EF \parallel DC$, which implies $K$ midpoint $CD$. Proof. First, we will prove that $EF \parallel DC$, which is true since $\measuredangle EDC = \measuredangle XDC = \measuredangle XFD = \measuredangle XEF = \measuredangle DEF$. Furthermore, by angle chasing, we can get that \[ \frac{1}{2} (\angle FDC + \angle ECK) = \angle FKE = \angle AFE = \angle ADC \]which suffices to prove what we wanted, and hence $K$ is the midpoint of $CD$. Claim 02. $AFXZ$ is cyclic. Proof. Notice that $FX$ is the radical axis of $\omega$ and $\Omega$, and therefore the perpendicular bisector of $FX$ passes through the center of both circle. Furthermore, $AB$ and $CD$ are tangents of $\omega$ and $\Omega$, which means that $AB, CD$, perpendicular bisector of $FX$ concur. Now, we claim that $AF = XZ$. Indeed, reflect wrt the perpendicular bisector of $FX$. Since $FX \parallel AZ$, we are done. This means that $AFXZ$ is an isosceles trapezoid, and hence, cyclic. Claim 03. $AFZC$ is cyclic. Proof. To prove this, it suffices to prove that $DF \cdot DA = DZ \cdot DC$. However, notice that $FM_1 \parallel AZ$, and therefore $\frac{DF}{DM_1} = \frac{DA}{DZ}$. Now, notice that \[ \frac{DA}{DZ} = \frac{DF}{DM_1} = \frac{DK}{DM_1} = 2 = \frac{DC}{DK} = \frac{DC}{DF} \]since $M_1$ has the same power wrt $\omega$ and $\Omega$.

Let $I$ be the center of the inscribed circle, and let $T=\overline{AB}\cap\overline{CD}$. Let $\gamma=(DFX)$, and let $A'$ and $D'$ be the reflections of $A$ and $D$ over $\overline{TI}$ respectively. Finally, let $\omega$ touch $\overline{CD}$ and ${AB}$ at $G$ and $H$. First, notice that $\omega$ and $\gamma$ are symmetric over $\overline{TI}$, since $\overline{AB}$ and $\overline{CD}$ are common external tangents of the two circles. This means that $\gamma$ is tangent to $\overline{AB}$ at $D'$, and $\overline{D'XA'}$ is tangent to $\omega$ at $X$. We now claim that $(D,G;A',C)=-1$. This is equivalent to showing that \[-1=(D,G;A'C)\stackrel{E}{=}(X,G;\overline{EA'}\cap\omega,E)\]which is true since $\overline{A'X}$ and $\overline{A'G}$ are tangent to $\omega$. Finally, we have \[\frac{\text{Pow}(A,\omega)}{\text{Pow}(A,\gamma)}=\frac{AH^2}{AD'^2}=\frac{A'G^2}{A'D^2}=\frac{CG^2}{CD^2}=\frac{\text{Pow}(C,\omega)}{\text{Pow}(C,\gamma)}\]so by the Forgotten Coaxality Lemma, we are done. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair G = dir(0); pair E = dir(95); pair F = dir(265); pair I = circumcenter(E,F,G); pair C = 2*circumcenter(I,E,G)-I; pair D = 2*circumcenter(I,F,G)-I; pair X = 2*foot(I,E,D)-E; pair T = extension(I,midpoint(F--X),C,D); pair D1 = 2*foot(D,I,T)-D; pair A = extension(F,D,D1,T); pair B = extension(E,C,D1,T); pair H = foot(I,A,B); pair A1 = 2*foot(A,I,T)-A; draw(A--B--C--D--cycle, black+1); draw(A--T); draw(D--T); draw(D1--A1); draw(E--A1); draw(E--D); draw(circumcircle(A,F,X), mydash); draw(circumcircle(E,F,G)); draw(circumcircle(D,F,X)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(90)); dot("$F$", F, dir(315)); dot("$G$", G, dir(0)); dot("$H$", H, dir(225)); dot("$T$", T, dir(270)); dot("$A'$", A1, dir(0)); dot("$D'$", D1, dir(225)); dot("$X$", X, dir(0)); dot("$I$", I, dir(0)); [/asy][/asy]

Here's My solution: Claim . $EF||CD$ Proof . Note that $\angle FEX = \angle DFX = \angle EDC$ so the claim holds . Now , let $AD \cap BC = Y$ .Note that $\omega$ is the incircle of $YDC$ . therefore $YF = YE$ , but $EF||CD$ so $\triangle YDC$ is an isosceles triangle . Now , to proving that the circles $\omega , (DFX) , (AXC)$ are coaxial , by coaxiality lemma , it's enough to prove that : $\frac{\text{Pow}(A,\omega)}{\text{Pow}(A,(DFX))}=\frac{\text{Pow}(C,\omega)}{\text{Pow}(C,(DFX))}$ Let $M$ be the midpoint of $CD$ we have: $\frac{\text{Pow}(C,\omega)}{\text{Pow}(C,(DFX))}$ $=$ $\frac {CM^2}{CD^2}$ $=$ $\frac {1}{4}$ So it's enough to prove that $\frac{\text{Pow}(A,\omega)}{\text{Pow}(A,(DFX))}$ $ = \frac {AF^2}{AF . AD}$ $= \frac{AF}{AD}$ $=$ $\frac {1}{4}$

and we're done .

Let the incircle meet $CD$ at $R$ and $AB$ at $P$. Let $H$ point where $AB$ is tangent to $(DXF)$. Some angle chasing yields \begin{align*} \measuredangle CDF=\measuredangle DXF=\measuredangle EXF=\measuredangle EFA=\measuredangle BEF=\measuredangle CEF, \end{align*}thus $DFEC$ is an isosceles triangle. Hence, $CR=CE=DF=DR$, which means that $R$ is the midpoint of $CD$. Note that $DF=DR=PH=AP+AH=AF+\sqrt{AF\cdot (AF+DF)}$, which simplifies to $DF=3AF$, hence $AH=2AP$. Now, we have easy finish by the coaxiality lemma, $$\frac{P(A,(DXF))}{P(A,(ERFP)))}=\frac{AH^2}{AP^2}=4=\frac{CD^2}{CR^2}=\frac{P(C,(DXF))}{P(C,(ERFP)))},$$we are done. [asy][asy]import geometry; size(12cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,E,F,D,C,B,A,R,X,H,P; O=(0,0);D=dir(160);C=dir(330);R=midpoint(C--D); F=intersectionpoints(circle(D,abs(D-R)),circle(O,1))[0];E=intersectionpoints(circle(C,abs(C-R)),circle(O,1))[0]; A=(4F-D)/3;path w=circumcircle(R,E,F);P=2foot(F,A,circumcenter(R,E,F))-F;B=intersectionpoint(line(C,E),line(A,P)); X=intersectionpoints(w,D--E)[0];H=extension(extension(P,R,D,E),C,A,B); path x=circumcircle(D,X,F); draw(w,heavyblue);draw(x,heavyblue); draw(A--B--C--D--cycle,deep);draw(H--A,deep);draw(circumcircle(A,F,C),heavyblue+dashed); draw(D--E,deep);draw(R--P,deep);draw(D--H,deep); dot("$E$",E,dir(E)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$C$",C,dir(C)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$R$",R,dir(R)); dot("$X$",X,dir(X)); dot("$H$",H,dir(H)); dot("$P$",P,dir(P)); [/asy][/asy]

Let circle DXF meet AB at S. Let SX meet CD at K. both circles are tangent to AB and CD so AD and SK are symmetric about Line passing through centers so SK is tangent to W at X. we will prove AFXKC is cyclic. step1 : EFDC is cyclic. ∠BEF = ∠EXF = 180 - ∠FXD = ∠FDC ---> EFDC is cyclic. step2 : AFXK is cyclic. AF and KX are symmetric about line passing through centers so AFXK is isosceles trapezoid. step3 : FXKC is cyclic. ∠FCD = ∠FED = ∠FXS ---> FXKC is cyclic. now we have both AFXK and FXKC are cyclic so AFXKC is cyclic as wanted. we're Done.

Trig spam, anyone? Also, first time using Forgotten Coaxiality Lemma thnx to Kagebaka's handout on POP which helped me recognize the picture Let $G$ be the other touchpoint of $(DFX)$ with $\overline{AB}$ and let $P$,$Q$ be the touchpoints of $\omega$ with $\overline{AB}$, $\overline{CD}$ respectively. $\measuredangle PGD = \measuredangle PGF + \measuredangle FGD = \measuredangle GDF + \measuredangle FDQ = \measuredangle GDQ$, thus $GPQD$ is an isosceles trapezium (better seen by taking the intersection of $AB$ and $CD$ and applying POP). By Radax on $\{(GFXD), \omega , (PFXQ)\}$ we have $XF \parallel GD \parallel PQ$, thus $FGDX$ and $FPQX$ are also isosceles trapeziums, moreover $\color{red}\triangle FGP \cong \triangle XDQ$. $\measuredangle EXF = \measuredangle DXF = \measuredangle DGF$ and $\measuredangle FEX = \measuredangle DFX = \measuredangle FDG$ which means $\color{blue}\triangle FDG \sim \triangle FEX$. Finally, \begin{align*} \dfrac{\text{Pow}(A, (PFXQ))}{\text{Pow}(A , (GFXD))} &= \dfrac{AF^2}{AG^2} \\ &= \dfrac{\sin^2 \angle AGF}{\sin^2 \angle AFG} \\ &\stackrel{\color{red}\cong ~ \triangle}{=} \dfrac{\sin^2 \angle XDQ}{\sin^2 \angle DFG} \\ &\stackrel{\color{blue}\sim ~ \triangle}{=} \dfrac{\sin^2 \angle EDC}{\sin^2 \angle EFX} \\ &= \dfrac{\sin^2 \angle EDC}{\sin^2 \angle CED} \\ &= \dfrac{CE^2}{CD^2} = \dfrac{CQ^2}{CD^2} = \dfrac{\text{Pow}(C, (PFXQ))}{\text{Pow}(C , (GFXD))} \end{align*}which, by the Forgotten Coaxiality Lemma, means that $AFXC$ is cyclic.

Define $\gamma = (DFX)$, $Q = \omega \cap CD$, and $Z$ as the intersection of the tangent to $\omega$ at $X$ and $CD$. We first notice that $Z$ is symmetric to $A$ about the line connecting the centers of $\omega$ and $\gamma$, as $F$, $X$ and lines $AB$, $CD$ are symmetric. We then perform the length chase: \[-1 = (XC \cap \omega, X; Q, E) \overset{X}{=} (CZ;QD) \implies \frac{CQ}{CD} = \frac{ZQ}{ZD}\]\begin{align*} &\implies CQ \cdot ZD = ZQ(QD+QC) \\ &\implies CQ(ZD-ZQ) = QD \cdot ZQ \\ &\implies CQ^2 + CQ \cdot ZQ = CQ^2 + ZD \cdot CQ - QD \cdot CD \\ &\implies \frac{CQ}{CQ+QD} = \frac{CQ-ZQ}{CQ+ZQ} = \frac{ZQ}{QD-ZQ} \\ &\implies \frac{CQ}{CD} = \frac{ZQ}{ZD}. \end{align*} We finish using Coaxiality Lemma, since this final equality implies \[\frac{\operatorname{pow}(C,\omega)}{\operatorname{pow}(C,\gamma)} = \frac{\operatorname{pow}(Z,\omega)}{\operatorname{pow}(Z,\gamma)} = \frac{\operatorname{pow}(A,\omega)}{\operatorname{pow}(A,\gamma)}. \quad \blacksquare\]

Claim: $EF\parallel CD$. We have $$\angle XDC=\angle XFD=\angle XEF$$by the given tangencies. In particular, since $EC,CD,DF$ are tangent to $\omega$, this means that $EFDC$ is an isosceles trapezoid by symmetry. Let $P$ be the tangency point of $(DXF)$ with $AB$, and let $Q$ and $R$ be the tangency points of $\omega$ with $AB$ and $CD$ respectively. We will use the Forgotten Coaxiality Lemma on $\omega$ and $(PFXD)$. For $C$, since due to the isosceles trapezoid $R$ is the midpoint of $CD$, we have $$CD=2CR=2CE,$$so the power of $C$ with respect to $(PFXD)$ is $4$ times its power with respect to $\omega$. For $A$, let let $AQ=AF=1$, and let $FD=DR=s$. Then, $AP=\sqrt{AF\cdot AD}=\sqrt{s+1}$. Furthermore, $PQ=RD$, so $1+\sqrt{s+1}=s$. Thus, $s=2$, so $AP=2AQ$, so the power ratio for $A$ is also $4$, done.

f=1/2π√LC