Lemma: Let $ABCD$ be a cyclic quadrilateral with circumcenter $O.$ Denote $E \equiv AC \cap BD$ and $F \equiv AB \cap CD.$ Let $F'$ be the antipode of $F$ on $\odot(FBC).$ Then $O, E, F'$ are collinear. 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dot((1014.2460788614324,1012.8210610116407),linewidth(3.pt) + dotstyle); label("$O$", (1014.3696252938954,1011.5795794101209), E * labelscalefactor); dot((1008.4015074645152,1007.0321130227279),linewidth(3.pt) + dotstyle); label("$F'$", (1008.0045583958155,1005.5054869988102), E * labelscalefactor); dot((1046.2228815301826,1011.1177047633228),linewidth(3.pt) + dotstyle); label("$G$", (1046.7041651361415,1010.0519633545817), NE * labelscalefactor); dot((1029.535728372191,1027.9651973007983),linewidth(3.pt) + dotstyle); label("$M$", (1029.6821576601335,1028.2014969668219), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Denote $G \equiv AD \cap BC$ and let $M$ be the Miquel point of $ABCD.$ It is known that $O, E, M$ are collinear, and $OM \perp FG.$ On the other hand, since $M \in \odot(FBC)$ and $\overline{FF'}$ is a diameter of this circle, we have $F'M \perp FG.$ Therefore, $O, E, F'$ are collinear. $\blacksquare$ __________________________________________________________________________________________________________________________________________________ Main Proof: Let $I$ be the incenter of $\triangle ABC.$ Note that $AI$ is the perpendicular bisector of $\overline{DE}$; also, $F$ is the reflection of $D$ in $IB$ and $G$ is the reflection of $E$ in $IC.$ It follows that $IF = ID = IE = IG.$ Therefore, $DEGF$ is cyclic with circumcenter $I.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); 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 = 983.8150410180909, xmax = 1049.4664952960939, ymin = 997.8046374990395, ymax = 1031.1689818563664; /* image dimensions */ pen fsfsfs = rgb(0.9490196078431372,0.9490196078431372,0.9490196078431372); pen evfuff = rgb(0.8980392156862745,0.9568627450980393,1.); pen qqzzff = rgb(0.,0.6,1.); filldraw((1002.3155835435821,1028.2479023843553)--(997.5658551273822,1009.082914299408)--(1021.4141418604702,1009.1493440674667)--cycle, fsfsfs, linewidth(0.8) + gray); filldraw((1007.5274341255129,1000.0825257520867)--(1014.231628649713,1016.3318572782222)--(998.261773264589,1011.8909196502241)--cycle, evfuff, linewidth(0.8) + qqzzff); 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label("$I$", (1005.9581933269889,1015.7585447765248), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $DF$ and $EG$ meet at $S.$ By the above observations, the sidelines of $\triangle SED$ are parallel to the sidelines of the intouch triangle of $\triangle ABC.$ Therefore, homothety carrying the incircle into $\omega$ carries the intouch triangle of $\triangle ABC$ into $\triangle SED.$ By homothety, we see that $S$ lies on major arc $\widehat{DE}$ of $\omega$, and the tangent to $\omega$ at $S$ is parallel to $BC.$ Therefore, $S$ and $L$ are antipodal points on $\omega.$ Applying the lemma to cyclic quadrilateral $DEGF$, we see that $I, K, L$ are collinear. $\square$

mofumofu wrote: Circle $\omega$ is tangent to sides $AB$,$AC$ of triangle $ABC$ at $D$,$E$ respectively, such that $D\neq B$, $E\neq C$ and $BD+CE<BC$. $F$,$G$ lies on $BC$ such that $BF=BD$, $CG=CE$. Let $DG$ and $EF$ meet at $K$. $L$ lies on minor arc $DE$ of $\omega$, such that the tangent of $L$ to $\omega$ is parallel to $BC$. Prove that the incenter of $\triangle ABC$ lies on $KL$. Suppose $DE$ is at variable distance $\lambda$ from $A$. Define $F, G$ as the reflections of $D, E$ in lines $BI, CI$ respectively. Let $S=DF \cap EG$ and $T=DE \cap BC$. Let $A_1, A_2$ be the touch-points of the incircle and $A$-excircle with $BC$. Observe that $\angle DFB=\angle DEG=90^{\circ}-\tfrac{1}{2}\angle B$ hence $DEFG$ is cyclic. Also $I$ is the center of the circle $\odot(DEFG)$. Now by Brokard's Theorem, we know $IK \perp ST$. Note that $DLES$ is cyclic and is the image of the incircle under suitable homothety at $A$. Accordingly, we see $S \in AA_1, L \in AA_2$ and $T \in BC$ move with constant velocity. Hence we can find a spiral similarity with pivot $P$ such that $BC \mapsto AA_2$ and $T \mapsto S$. Let $Z$ be point at infinity with $ZP \perp ST$. Note that $L \mapsto Z$ is projective. Thus, in order to prove $I, K, L$ collinear, it is equivalent to show $L \mapsto Z$ is a perspectivity about $I$. Hence it is enough to verify the result for three values of $\lambda \in \mathbb{R}$. If $DE$ is an intouch chord then we're done. Suppose $B=D$ (likewise let $C=E$ for the final case). Then $K=B$ and $\angle LDE=\tfrac{1}{2}\angle C$ hence $L$ lies on $BI$ and we're done. $\blacksquare$

Too easy for China TST P3. I will provide just a sketch since I am on mobile. By easy angle chasing, $DEFG$ is cyclic and the intersection $DF\cap EG$ lies on $\omega$. By Reim, $LP$ is diameter of $\omega$. Note that $I$ is the center of $\odot(DEFG)$. Now add Miquel point $M$ of $DEFG$ which lies on $IK$ and $\omega$. Note that $\angle LMP=90^{\circ}$ hence $I, K,L,M$ are colinear and we are done.

It is also susceptible to straight barycentric coordinates. Substitute $a=BC=v+w$, $b=CA=w+u$, $c=AB=u+v$. Let $X=(0:w:v)$, $Y=(w:0:u)$, $Z=(v:u:0)$ be the contact points of the incircle on $\overline{BC}$. Then we may set \begin{align*} D &= (v-t:u+t:0) \\ E &= (w-t:0:u+t) \\ F &= (0:w+t:v-t) \\ G &= (0:w-t:v+t) \end{align*}where $t=XF=XG=EY=DZ$. We now solve for $K = (t:y:z)$, say. Since $K = \overline{DG} \cap \overline{EF}$ the standard determinants give \begin{align*} t(u+t)(v+t) &= (v-t)\left[ (v+t)y-(w-t)z \right] \\ t(u+t)(w+t) &= (w-t)\left[ -(v+t)y+(w+t)z \right]. \\ \implies t(u+t)\left[ (v+t)(w-t)+(w+t)(v+t) \right] &= \left[ -(w-t)^2(v-t)+(v+t)(w-t)(w+t) \right]z \\ z &= \frac{t(u+t)(v+t)(2w)} {(w-t)\left[ (v+t)(w+t)-(w-t)(v-t) \right]} \\ &= \frac{w(u+t)(v+t)}{(w-t)(v+w)}. \end{align*}Thus we recover \[ K = \left( \frac{t(v+t)}{u+t} : \frac{v(w+t)}{v-t} : \frac{w(v+t)}{w-t} \right). \]On the other hand, by homothety we have \begin{align*} \vec L &= \left( 1 + \frac tu \right) [2 \vec I - \vec X] - \frac tu \vec A \\ \left( 2u+2t \right) \vec I - u \vec L &= \left( u+t \right) \vec X + t \vec A \\ &= t \vec A + (u+t) \frac{w \vec B + v \vec C}{w+v} \\ \end{align*}Consequently the point $W = \frac{(2u+2t) \vec I - u \vec L}{u+2t} = \left( t(v+w) : w(u+t) : v(u+t) \right)$ lies on $\overline{IL}$ and is distinct from $I$ (since $u \neq 0$). Thus it is enough to show $W$, $K$, $I$ collinear: \begin{align*} \det \begin{bmatrix} \frac{t(v+w)}{u+t} & \frac{v(w+t)}{v-t} & \frac{w(v+t)}{w-t} \\ v+w & w+u & u+v \\ t(v+w) & w(u+t) & v(u+t) \end{bmatrix} &= \det \begin{bmatrix} 0 & \frac{v(w+t)}{v-t}-w & \frac{w(v+t)}{w-t}-v \\ v+w & w+u & u+v \\ t(v+w) & w(u+t) & v(u+t) \end{bmatrix} \\ &= \det \begin{bmatrix} 0 & \frac{v(w+t)}{v-t}-w & \frac{w(v+t)}{w-t}-v \\ v+w & w+u & u+v \\ 0& u(w-t) & u(v-t) \end{bmatrix} \\ &= -(v+w) \det \begin{bmatrix} \frac{t(v+w)}{v-t} & \frac{t(v+w)}{w-t} \\ u(w-t) & u(v-t) \end{bmatrix} = 0. \end{align*}

My proof : From simple angle chasing one can easily get that $\angle ((EDGF), (EDK)) = \angle ((EDGF), \omega)$. So in other words $(EDK)$ is inversion image of $\omega$ wrt $(EDGF)$, so $I$ is the inner homothety center of $\pi, (EDK)$. Now note that the tangent to $(EDK)$ from $K$ is parallel to $BC$, so from a homothety we get that $K, L, I$ are collinear. Done

Solution. Let $X=DF \cap EG$. Since $DEX$ and the intouch triangle of $ABC$ are homothetic, $X \in \omega$. Shooting lemma for $\omega$ and point $X$ implies $FGED$ is cyclic; its center is the intersection of the $\perp$ bisectors of $DF, EG$, i.e., $I$. Draw in the Miquel point of $FGED$, since it is the inverse of $K$ in $FGED$ and also the foot of $I$ the polar of $K$ wrt this circle, we get that the antipode of $X$ ($L$), $I$, $K$ are collinear. Done.

Let $ Y, Z $ be the second intersection of $ \omega $ with $ EF, DG, $ respectively. Obviously, the incenter $ I $ of $ \triangle ABC $ is the circumcenter of $ DEFG, $ so simple angle chasing yields $ \measuredangle IFG = \measuredangle LYZ, $ $ \measuredangle IGF = \measuredangle LZY. $ On the other hand, by Reim's theorem we get $ FG \parallel YZ, $ so $ \triangle IFG $ and $ \triangle LYZ $ are homothetic $ \Longrightarrow I, L, K $ are collinear.

You can solve this question easily if using Simson theorm and what you need to do is to find the obvious circle of DEFG and prove two right angles.Maybe a little not natural but quite simple.China`s D1 questions are often easy.Then the whole solving is your task and I just offer an idea

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It is easily to solve just by Pascal's theorem with circle DEFG. We could find that the incenter I of triangle ABC is the circumcenter of DEFG, and that make it natual to use Pascal's theorem.

My solution: Let $I$ be the incenter of $\triangle ABC$. Also let $DF \cap EG=T,\odot (FGT) \cap \omega=P,DE \cap BC=X$. Note that, as $AD=AE,BD=BF,CE=CG$, we have that $I$ lies on the perpendicular bisectors of $DE,DF,CG$, which in turn gives $ID=IE=IF=IG$. This means that $D,E,F,G$ lie on a circle centered at $I$. Also, $\angle BDT=\angle BFD=\angle DET \Rightarrow T$ lies on $\omega$, as $BD$ is tangent to $\omega$. Now, $DE$ is antiparallel to $FG$ as well as to the tangent to $\omega$ at $T$, which means that the tangent to $\omega$ at $T$ is parallel to $BC$. This also gives that $L$ is the antipode of $T$ in $\omega$, and so $\measuredangle LPT=90^{\circ}$. But, $P$ is the Miquel Point of the degenerate cyclic quadrilateral $DEFG$, and so, by Brocard's Theorem, we get that $I$ is the orthocenter of $\triangle KTX$, and that $P$ lies on $TX$. Thus, $\measuredangle IPT=90^{\circ}=\measuredangle LPT \Rightarrow P,K,I,L$ are collinear. $\blacksquare$

Different. Let $I$ be the incenter of $\triangle ABC$; $F_1$, $G_1$ are reflections of $F$, $G$ in $I$. Clearly, $I$ lies on perpendicular bisector of $DE$,$DF$,$EG$; so $DEFGF_1G_1$ is cyclic and center is $I$. Claim: $D$, $L$, $F_1$ and $E$, $L$, $G_1$ are collinear. Proof: I will prove that $\angle LDE= \angle F'DE$, which gives the first collinearity; second one follows similarly. $$\angle LDE= \angle AEL=\angle (HE,BC)=\frac{1}2 \angle C$$$$\angle F'DE=\angle F'FE=90-\angle FF'E=90-\angle EGC=\frac{1}2 \angle C$$This completes the proof of claim. $\square$ Now, by Pascal's Theorem on $FF'DGG'E$, we get $K,I,L$ are collinear. $\square$

[asy][asy] size(5cm); defaultpen(fontsize(10pt)); pen pri=deepblue; pen sec=deepcyan; pen tri=deepgreen; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,I,D,EE,F,G,K,X,L,M; A=dir(110); B=dir(210); C=dir(330); I=incenter(A,B,C); D=(A+4B)/5; EE=reflect(A,I)*D; F=reflect(B,I)*D; G=reflect(C,I)*EE; K=extension(D,G,EE,F); X=extension(D,F,EE,G); L=extension(K,I,X,X+(0,1)); M=foot(X,I,K); filldraw(circumcircle(D,F,G),tfil,tri); filldraw(circumcircle(D,EE,L),sfil,sec); filldraw(incircle(A,B,C),fil,pri); filldraw(D--X--EE--cycle,sfil,sec); draw(L--M--X,tri); draw(D--G,sec); draw(EE--F,sec); filldraw(A--B--C--cycle,fil,pri); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,W); dot("$I$",I,W); dot("$E$",EE,dir(30)); dot("$F$",F,SW); dot("$G$",G,SE); dot("$K$",K,dir(110)); dot("$X$",X,S); dot("$L$",L,N); dot("$M$",M,dir(240)); [/asy][/asy] Let $X=\overline{DF}\cap\overline{EG}$ and $M$ be the Miquel point of $DFGE$. Note that $\triangle XED$ and the contact triangle are homothetic at $A$, so $\overline{XL}$ is a diameter of $\omega$. Since $ID=IE$, $ID=IF$, $IE=IG$ by construction, $I$ is the circumcenter of cyclic quadrilateral $DFGE$, thus by properties of the Miquel point, $I$, $K$, $M$ lie on a line perpendicular to $\overline{XM}$. But $\angle XML=90^\circ$, so $L$, $I$, $K$, $M$ are collinear, as desired

We proceed with barycentric coordinates. Let $AD=AE=r$. Then, the coordinates of $F,G$ are $(0,a-c+r,c-r)$ and $(0,b-r,a-b+r)$ respectively, and the coordinates of $D,E$ are $(c-r,r,0)$, $(b-r,0,r)$ respectively. Computing a determinant, the equations of $EF$ and $DG$ are $$(a-b+r)rx-(a-b+r)(c-r)y+(b-r)(c-r)z=0$$$$(a-c+r)x+(b-r)(c-r)y-(a-c+r)(b-r)z=0$$Solving, we get that point $K$ has coordinates $$((a^2+2ar-ab-ac)(b-r)(c-r),r(b-r)(a+c-b)(a-c+r),r(c-r)(a+b-c)(a-b+r))$$Also, if we consider point $X=(0,s-b,s-c)$, the $A$-extouch point, note that $L$ is a homothety of $X$ wrt $A$ with factor $\frac{r}{s}$. So, it has coordinates $((s-r)a,r(s-b),r(s-c))$. Now, if we put the coordinates of $I,K,L$ into a determinant, we get a cubic in $r$, since it is easily verified that the quartic coefficient equals $0$. Thus, we just need to verify that the determinant is $0$ at $4$ points. We will choose $0,b,c,s-a$. At $0$, we have $$\begin{vmatrix}a&b&c\\(a^2-ab-ac)bc&0&0\\as&0&0\end{vmatrix}=0$$At $r=b$ we have $$\begin{vmatrix}a&b&c\\0&0&ab(c-b)(a+b-c)\\a(s-b)&b(s-b)&b(s-c)\end{vmatrix}=ab(c-b)(a+b-c)(ab(s-b)-ab(s-b))=0$$and we get a similar result for $r=c$. Finally, at $r=s-a$, we have $$\begin{vmatrix}a&b&c\\0&2(s-a)(s-c)^2(s-b)&2(s-a)(s-b)(s-c)^2\\a^2&(s-a)(s-b)&(s-a)(s-c)\end{vmatrix}\propto\begin{vmatrix}a&b&c\\0&(s-c)&(s-b)\\a^2&(s-a)(s-b)&(s-a)(s-c)\end{vmatrix}$$$$=a(s-c)^2(s-a)-a(s-b)^2(s-a)+b(s-b)a^2-c(s-c)a^2=0$$as desired.

Maybe similiar to the others but since i like this ill post my solution anyways. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -17.64875782250771, xmax = 12.985228170215747, ymin = -10.078567823113621, ymax = 10.151798251301829; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.43637095285195,6.498396949998834)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4) + sexdts); draw((-5.43637095285195,6.498396949998834)--(6.96,-2.55), linewidth(0.4) + sexdts); draw((6.96,-2.55)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4) + sexdts); draw(circle((-2.756624790103522,-1.8758367417264656), 5.184106943530284), linewidth(0.4) + dtsfsf); draw(circle((-3.596878261132844,0.7499638003166852), 3.55859333554701), linewidth(0.4) + dtsfsf); draw((-3.596878261132844,0.7499638003166852)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4) + rvwvcq); draw((-3.596878261132844,0.7499638003166852)--(6.96,-2.55), linewidth(0.4) + rvwvcq); draw((-7.676150691333448,-0.24083530163939865)--(-1.283252160384008,-2.7527894567191105), linewidth(0.4)); draw((-5.735469886770201,-2.8623169582879795)--(0.2997740533078106,2.311452466335568), linewidth(0.4) + rvwvcq); draw((-3.684395533087758,4.307480807290673)--(-3.5093609891779307,-2.8075532066573032), linewidth(0.4)); draw((-7.676150691333448,-0.24083530163939865)--(-5.735469886770201,-2.8623169582879795), linewidth(0.4) + wvvxds); draw((0.2997740533078106,2.311452466335568)--(-1.283252160384008,-2.7527894567191105), linewidth(0.4) + rvwvcq); draw((-5.43637095285195,6.498396949998834)--(-2.8841186772775,3.306702222014393), linewidth(0.4) + wrwrwr); draw(circle((-3.5968782949371954,0.7499637752913503), 4.197873997018878), linewidth(0.4) + linetype("4 4") + rvwvcq); draw((-7.676150691333448,-0.24083530163939865)--(0.2997740533078106,2.311452466335568), linewidth(0.4)); draw((-2.6291309029295515,-7.0583757054673235)--(-16.712165924866934,-3.132350942791587), linewidth(0.4)); draw((-2.8841186772775,3.306702222014393)--(-5.546903657206548,-6.2449680793904125), linewidth(0.4)); draw((-5.735469886770201,-2.8623169582879795)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4) + wvvxds); draw((-1.283252160384008,-2.7527894567191105)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4) + wvvxds); draw((-5.43637095285195,6.498396949998834)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4)); draw((-2.8841186772775,3.306702222014393)--(-2.6291309029295515,-7.0583757054673235), linewidth(0.4)); draw((-16.712165924866934,-3.132350942791587)--(-7.676150691333448,-0.24083530163939865), linewidth(0.4)); draw((-16.712165924866934,-3.132350942791587)--(-8.57057841452274,-2.9320625037227908), linewidth(0.4)); /* dots and labels */ dot((-5.43637095285195,6.498396949998834),dotstyle); label("$A$", (-5.3553790280251805,6.690555678098006), NE * labelscalefactor); dot((-8.57057841452274,-2.9320625037227908),dotstyle); label("$B$", (-8.498346422083822,-2.738346504077927), NE * labelscalefactor); dot((6.96,-2.55),dotstyle); label("$C$", (7.037460759940218,-2.3603947288430267), NE * labelscalefactor); dot((-3.596878261132844,0.7499638003166852),linewidth(4pt) + dotstyle); label("$I$", (-3.5252967479404025,0.9019258573950597), NE * labelscalefactor); dot((-7.676150691333448,-0.24083530163939865),linewidth(4pt) + dotstyle); label("$D$", (-7.603197480738006,-0.07279187873705148), NE * labelscalefactor); dot((0.2997740533078106,2.311452466335568),linewidth(4pt) + dotstyle); label("$E$", (0.3735741965880374,2.4734095544243817), NE * labelscalefactor); dot((-3.5093609891779307,-2.8075532066573032),linewidth(4pt) + dotstyle); label("$T$", (-3.4258357544575344,-2.6388855105950584), NE * labelscalefactor); dot((-1.498834977839814,3.62429866411189),linewidth(4pt) + dotstyle); label("$M$", (-1.4167236861035932,3.786294668398246), NE * labelscalefactor); dot((-6.973851822223395,1.8722989086934119),linewidth(4pt) + dotstyle); label("$J$", (-6.887078327661354,2.0357811830997603), NE * labelscalefactor); dot((-5.735469886770201,-2.8623169582879795),linewidth(4pt) + dotstyle); label("$F$", (-5.653762008473786,-2.6985621066847796), NE * labelscalefactor); dot((-1.283252160384008,-2.7527894567191105),linewidth(4pt) + dotstyle); label("$G$", (-1.1979095004412827,-2.599101113201911), NE * labelscalefactor); dot((-3.684395533087758,4.307480807290673),linewidth(4pt) + dotstyle); label("$K$", (-3.604865542726697,4.462629424081752), NE * labelscalefactor); dot((-2.8841186772775,3.306702222014393),linewidth(4pt) + dotstyle); label("$L$", (-2.8091775948637503,3.468019489253067), NE * labelscalefactor); dot((-4.24854799053207,-1.5876392286151881),linewidth(4pt) + dotstyle); label("$K$", (-4.16184710623076,-1.425461390104063), NE * labelscalefactor); dot((-2.6291309029295515,-7.0583757054673235),linewidth(4pt) + dotstyle); label("$N$", (-2.5505790118082925,-6.8958160316618295), NE * labelscalefactor); dot((-16.712165924866934,-3.132350942791587),linewidth(4pt) + dotstyle); label("$O$", (-16.634255688982453,-2.9770528884368113), NE * labelscalefactor); dot((-5.546903657206548,-6.2449680793904125),linewidth(4pt) + dotstyle); label("$P$", (-5.474732220204623,-6.080235885102308), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Firstly we give a defination of few points. Let $DF \cap GE=N$ and $DE \cap FG=O$. Let $J,M,T$ be the touchpoints of the incircle with $AB,AC,BC$. Firstly notice that $\Delta DNE$ and $\Delta JTM$ are homothetic and so taking a homothety centred at $A$ s.t. $\odot(I) \mapsto \omega \implies N \in \omega$. Also $\{A,T,N\}$ is a collinear triplet $\implies L$ is the $N$-Antipode w.r.t $\omega$. Now let $P$ be the miquel point of quaderilaeral $DEGF$. Notice that $ID=IF=IG=IE \implies I$ is the centre of $\odot(DFGE)$. Now its well known that $K,P$ are inverses w.r.t $\odot(DEFG)\implies \{I,K,P\}$ is a collinear triplet. By brocard we have that $\angle IPN=90^\circ$. Now notice that since $L$ is the $N$-antipode $\implies \angle LPN=90^\circ \implies \{L,I,K,P\}$ is a collinear triplet as desired.$\blacksquare$.

Edit: Because the problem is a cyclic quad problem.

@above the fact that u can reduce ll the incenter stuff to focus just on the quad. because we kmow so much about miquel theory, it makes sense to focus on a structure we already know a lot about

Too easy for China TST P3. Anyways really Nice Problem. mofumofu wrote: Circle $\omega$ is tangent to sides $AB$,$AC$ of triangle $ABC$ at $D$,$E$ respectively, such that $D\neq B$, $E\neq C$ and $BD+CE<BC$. $F$,$G$ lies on $BC$ such that $BF=BD$, $CG=CE$. Let $DG$ and $EF$ meet at $K$. $L$ lies on minor arc $DE$ of $\omega$, such that the tangent of $L$ to $\omega$ is parallel to $BC$. Prove that the incenter of $\triangle ABC$ lies on $KL$. Note that after Simple Angle Chasing $DFGE$ we get that $DFGE$ is a cyclic quadrilateral. So, $\angle EGC=\angle GEC=\angle FDE$, now as $AC$ is tangent to $\omega$, hence it's obvious that $DF\cap EG=X\in\omega$. Now it's easy to notice that $L$ is the $K-\text{Antipode}$ WRT $\omega$. Let $LK\cap\omega=Y$, then it's well known that $Y$ is the Miquel Point of $DEGF$. So, the Circumcenter $(O)$ of $DEGF$ must lie on $LK$, and if $T$ is the Circumcenter of $\omega$ then $AT$ is the perpendicular bisector of $ED$. Hence, $ED\cap LY=O$. Hence, $OD=OF=OE=OG$. Hence, $BO,CO$ are the $B,C$ angle bisectors of $\triangle ABC$. So, $O\equiv I\in LK$. $\blacksquare$

Easy for a China #3 ? Interesting problem though.

Here is a moving points approach. First, by easy angle chasing we get $DF, EG$ concur at a point $M$ on $\omega$, and points $D,E,F,G$ all lie on a circle centered at $I$. Let $DE,BC$ intersect at $T$, and let $TM$ intersect $\omega$ again at $N$. By Brokard's theorem, $KI$ passes through $N$, so it suffices to prove that $LI$ also passes through $N$. This can be deduced from the following claim: Claim. Let $X = \mu D + (1-\mu)E$, $Y = \mu F + (1-\mu) G$ for some real number $\mu$. Denote the line through $X$ perpendicular to $DE$ by $\ell_1$, and denote the line through $Y$ perpendicular to $FG$ by $\ell_2$. Suppose $\ell_1 \cap \ell_2 = Z$. Then then as $\mu$ varies, $Z$ moves on the line $NL$ linearly. In particular, taking $\mu = 0.5$, we have $N,I,L$ collinear. Proof. Let $\phi_1$ denote the linear map that sends $X$ to the point $Z_1 = \ell_1 \cap NL$, and let $\phi_2$ denote the linear map that sends $Y$ to the point $Z_2 = \ell_2 \cap NL$. It suffices to verify that $\phi_1, \phi_2$ coincide for two distinct values of $\mu$: Take $\mu = 1$, then $\ell_1,\ell_2$ intersect on $NL$ because $T,D,F,N$ are concyclic. Take $\mu = 0$, then $\ell_1,\ell_2$ intersect on $NL$ because $T,E,G,N$ are concyclic. $\square$

Not the best solution, but... Bary wrt $\Delta ABC$. Let $AD=AE=t$ so that $\quad$-by lengths $D=(c-t:t:0)$, $E=(b-t:0:t)$, $F=(0:a-c+t:c-t)$, $G=(0:b-t:a-b+t)$. $\quad$-by a homothety from $A$ with factor $\frac{t}{s}$ sending the $A$-excircle touch point $(0:s-b:s-c)$ to $L$ we get $L=(a(s-t):t(s-b):t(s-c))$. Since $I=(a:b:c)$ it remains to compute $K$, the intersection of $$DG: t(a-b+t)x-(c-t)(a-b+t)y+(b-t)(c-t)z=0$$$$EF: t(a-c+t)x+(b-t)(c-t)y-(b-t)(a-c+t)z=0$$and solving we get $$K=(a(b-t)(c-t)(a+2t-b-c): t(a-c+t)(a-b+c)(b-t): t(a-b+t)(a+b-c)(c-t))$$Now to show the determinant $[IKL]=0$: -divide the first column by $a$. -multiply the first row by $t$, then divide the second and third columns by $t$ -add the first row to the second, then divide the second by $s$ -subtract $t$ times the second row from the first. Now we have $$\begin{pmatrix} 0 & b-t & c-t\\ 1 & 1 & 1\\ (b-t)(c-t)(a+2t-b-c) & (a-c+t)(a-b+c)(b-t) & (a-b+t)(a+b-c)(c-t) \end{pmatrix}$$ $$=(b-t)(c-t)(0)=0.$$where everything cancels (yay!)

We have $\angle (DF, EG) = 180 - \angle DFB - \angle EGC = 90 - \frac{\angle BAC}{2} = \angle ADE$, which is the measure of arc $DE$, so we have $X = DF \cap EG \in \omega$. So, $\angle BFD = \angle BDX = \angle DEG$ so $DFEG$ is cyclic. Note that since $I$ lies on the perpendicular bisectors of $EG, DF$, it must in fact be the circumcenter (since the quadrilateral is cyclic). Let $M$ be the miquel point of $DFEG$, which lies on $\omega$. It's well known that $I,K,M$ must be collinear, so it suffices to show that $MI$ goes through $L$. Since $IK \perp XM$, we must have that $MI \cap \omega$ is the antipode of $X$, so it suffices to show that the tangent at $X$ to $\omega$ is parallel to $BC$. But this is true since if $Z$ is a point on the tangent, then $\angle FXZ = \angle DXZ = \angle DEG = \angle DFB$, so we are done. $\blacksquare$

Let $I$ be incenter of $\triangle ABC$ Claim: $D,F,G,E$ lie on circle with center as $I$. $$\angle FDE = 180 - \angle EDA - \angle BDF = 90 - \frac{\angle C}{2} = \angle CGE$$hence $D,F,E,G$ cyclic.Note $I$ lie on perpendicular bisector of $DE$ and by condition $BD=BF$ and $CE=CG$ we have $I$ also lie on perpendicular bisector of $GF$ also from fact that $IB$ and $IC$ are angle bisector we have $I$ as center of $(DEF)$ Let $T=DF \cap GE$ Claim : $T$ lie on $(DLE)$ with $\angle TDL = \angle LET = 90$ Let perpendicular from $L$ to $BC$ meet at $H$.If $O$ is center of $(DLE)$ then $O,E,B,H$ cyclic. $$\angle LOD = \angle CBA \implies \angle ODL = 90 - \frac{\angle B}{2} \implies \angle EDL = \frac{\angle C}{2}$$hence $\angle FDL = \angle FDE + \angle EDL = 90$ Claim : $L,I,K$ are collinear. Let $M=(TGF)\cap (DTE)$. Then $M$ is miquel point point of $\Box DEGF$. hence $\overline{I-K-M}$ (well know property) as $LT$ is diameter $\angle TML = 90$, but it's well know that $\angle TMI = 90$ hence $\overline{L-I-M}$ Hence $L,I,K$ are collinear $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(19.07651842087848cm); 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 = -4.322575073781325, xmax = 22.753943347097152, ymin = -11.155828560285837, ymax = 3.1133695277123317; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); draw((6.993497304868754,1.7930527604426043)--(4.74121400092014,-6.783937749194878)--(15.721214000920138,-6.883937749194877)--cycle, linewidth(0.4) + ffxfqq); /* draw figures */ draw((6.993497304868754,1.7930527604426043)--(4.74121400092014,-6.783937749194878), linewidth(0.4) + ffxfqq); draw((4.74121400092014,-6.783937749194878)--(15.721214000920138,-6.883937749194877), linewidth(0.4) + ffxfqq); draw((15.721214000920138,-6.883937749194877)--(6.993497304868754,1.7930527604426043), linewidth(0.4) + ffxfqq); draw(circle((10.260220799534883,-3.6489912612960405), 6.347229752037104), linewidth(0.4)); draw((xmin, -3.673826499279709*xmin + 27.485948481710658)--(xmax, -3.673826499279709*xmax + 27.485948481710658), linewidth(0.4)); /* line */ draw(circle((8.962592300056919,-5.441060612478726), 3.7418809170599285), linewidth(0.4) + red); draw((5.343414536894738,-4.490679049058201)--(9.911189132387893,-6.831023132905878), linewidth(0.4)); draw((7.11212409920384,-6.805530737339539)--(11.600778937775456,-2.7874509836861616), linewidth(0.4)); draw((xmin, -0.009107468123861607*xmin-1.6173979879750853)--(xmax, -0.009107468123861607*xmax-1.6173979879750853), linewidth(0.4)); /* line */ draw((xmin, 4.764129681840161*xmin-44.56063720954149)--(xmax, 4.764129681840161*xmax-44.56063720954149), linewidth(0.4) + green); 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dot((9.911189132387893,-6.831023132905878),linewidth(4.pt) + dotstyle); label("$G$", (9.962512989815947,-6.023366441774246), NE * labelscalefactor); dot((7.11212409920384,-6.805530737339539),linewidth(4.pt) + dotstyle); label("$F$", (6.879857723589172,-6.372945904954602), NE * labelscalefactor); dot((8.996669947959758,-1.6993348727470368),linewidth(4.pt) + dotstyle); label("$L$", (9.311023990252556,-1.4152735180331897), NE * labelscalefactor); dot((8.538392964454843,-3.882625852366973),linewidth(4.pt) + dotstyle); label("$I$", (8.738984868684701,-3.957669613890324), NE * labelscalefactor); dot((8.112914677045005,-5.9096595903947025),linewidth(4.pt) + dotstyle); label("$K$", (8.500635234698095,-5.880356661382282), NE * labelscalefactor); dot((8.928514652154083,-9.182786352210424),linewidth(4.pt) + dotstyle); label("$T$", (9.199794161058806,-9.598610951573342), NE * labelscalefactor); dot((10.202416004181245,-9.995957791125791),linewidth(4.pt) + dotstyle); label("$A'$", (10.264422526198981,-9.868740536758162), NE * labelscalefactor); dot((7.489289618033217,-8.880690244372106),linewidth(4.pt) + dotstyle); label("$M$", (7.213547211170421,-8.45453270843763), NE * labelscalefactor); dot((-2.8283481155758152,-6.714998203507301),linewidth(4.pt) + dotstyle); label("$R$", (-2.76535746506883,-6.595405563342101), NE * labelscalefactor); dot((8.950012986401948,-6.822269251794895),linewidth(4.pt) + dotstyle); label("$H$", (8.389405405504345,-6.54773563654478), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Finally done after and Let $DF\cap EG=J$ and $DE\cap FG=S$, we have $\odot(DEGF)$ with center $I$. Note that the $J$ is $L$-antipode this is because $\triangle DEJ$ and contact triangle are homothetic with center $A$, so $\overline{A-J-N}$. Now let the miquel point of $\odot(DEGF)$ be $M$ which is also $\odot(DJE)\cap \odot(FJG)$, then $\overline{I-K-M}$ and finally by Brokard's theorem we have $\angle IMJ=90^{\circ}$ because $J$ is the $L$-antipode we have $\angle LMJ=90^{\circ}$, so $\overline{L-I-K-M}$ and we are done!

Here is an alternative solution using my favorite technique, the ``radical axis trick'': if you don't know how to prove three points are collinear, try to throw them on a radical axis. While there are certainly more natural solutions (see all above posts), this problem is still a nice showcase of the method. Claim: Points $D$, $E$, $F$, and $G$ lie on a circle with center $I$. Proof. From the given length conditions, we immediately deduce that $ID = IF$ and $IE = IG$. But we also have $ID = IE$ from the definition of $\omega$. $\blacksquare$ [asy][asy] size(300); defaultpen(fontsize(10pt)); pair A, B, C, D, E, F, G, I, O, K, L, X, Y, MA; real r = 0.47; A = dir(120); B = dir(210); C = dir(330); I = incenter(A, B, C); O = (1+r)*I - r*A; D = foot(O, A, B); E = foot(O, A, C); F = 2*foot(D, B, I) - D; G = 2*foot(E, C, I) - E; K = extension(D, G, E, F); L = O + abs(O-D)*dir(90); MA = dir(270); X = 2*foot(MA, D, L) - D; Y = 2*foot(MA, E, L) - E; draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(CP(O, D), heavyred); draw(D--G^^E--F, lightblue); draw(circumcircle(B, D, I), lightblue+dashed); draw(circumcircle(C, E, I), lightblue+dashed); draw(D--L--Y, heavycyan); draw(circumcircle(D, F, G), dotted+heavygreen); draw(K--L, dotted+magenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$D$", D, dir(130)); dot("$E$", E, dir(50)); dot("$I$", I, dir(120)); dot("$F$", F, dir(300)); dot("$G$", G, dir(315)); dot("$K$", K, dir(270)); dot("$L$", L, dir(90)); dot("$X$", X, dir(90)); dot("$Y$", Y, dir(60)); dot("$M_A$", MA, dir(270)); [/asy][/asy] Remark: If we are looking to apply the radical axis trick, we now know that $K$ automatically lies on the radical axis of any two circles $\omega_1$ and $\omega_2$ with $D, G\in \omega_1$ and $E, F\in \omega_2$. So there are two natural choices: $\omega_1 = (DGI)$ and $\omega_2 = (EFI)$, or $\omega_1 = (DGL)$ and $\omega_2 = (EFL)$. Trying both briefly should convince you that the first option is much easier. At this point, we make the following observations: Since $DI = IG$ and $\overline{BI}$ bisects $\angle DBG$, $B\in (DGI)$. Similarly, $C\in (EFI)$. (This is another of my favorite tricks.) Straightforward angle chasing yields $\overline{DL}\parallel \overline{BI}$ and $\overline{EL}\parallel \overline{CI}$. (For example, the homothety at $A$ sending $\omega$ to the $A$-excircle sends $L$, $D$, $E$ to the excircle tangency points.) To finish, let $X = \overline{DL}\cap (DGI)\neq D$ and $Y = \overline{EK}\cap (EFI)\neq E$. It suffices to show that $DXEY$ is cyclic; then $K$, $I$, $L$ will all lie on the radical axis of $(DGI)$ and $(EFI)$. We will show that $D$, $X$, $E$, $Y$ all lie on a circle with center $M_A$, the midpoint of arc $BC$. First note that $M_AD = M_AE$, by the definition of $\omega$. Furthermore, from the parallel lines, we see that $BDXI$ and $CYEI$ are isosceles trapezoids. Hence $DX$ and $BI$ share a perpendicular bisector, and so if $M_A$ is the midpoint of arc $BC$, we must have $M_AD = M_AX$. Similarly, we get $M_AE = M_AY$, which completes the proof.