If the incircle $(I)$ of $\triangle ABC$ touches $AC,AB,BC$ at $X,Y,Z,$ from the homothetic triangles $\triangle NMC \sim \triangle BAC$ and $\triangle QPB \sim \triangle CAB,$ it follows that $NE \parallel BX$ and $QF \parallel CY.$ Together with $AE=EF,$ it follows that figures $\triangle AXY \cup BX \cup CY$ and $\triangle AEF \cup NE \cap QF$ are homothetic $\Longrightarrow$ $\triangle ABC$ and $\triangle ARS$ are homothetic. It remains to show that the incenter $U$ of $\triangle AXY$ lies on $(I),$ which is quite direct, because of $\angle XUY= 90^{\circ}+\frac{_1}{^2}\angle BAC$ and $\angle XZY= 90^{\circ}-\frac{_1}{^2}\angle BAC$ $\Longrightarrow$ $UXZY$ is cyclic. Consequently, in the homothetic figure, the incenter of $\triangle AEF$ lies on the incircle of $\triangle ARS.$

We'll prove $RS//BC$ +Let $G;K$ is tangency point of the incircle $\triangle MNC;\triangle PQB$ with $MN;PQ$ respectively.It's easy to see $F;G;K;E$ are collinear.Morever $\dfrac{PK}{KQ} =\dfrac{ME}{EC} \Rightarrow PM;EF;BC$ concurrent at $L$ +Applying Menelaus's theorem.We'll get: - $\dfrac{EC}{EA}.\dfrac{NB}{NC}.\dfrac{RA}{RB}=1$ - $\dfrac{FA}{FB}.\dfrac{QC}{QB}.\dfrac{SA}{SC}=1$ +To prove $RS//BC$,We'll prove $\dfrac{RA}{RB}=\dfrac{SA}{SC}$ or $\dfrac{NB}{NC}.\dfrac{QB}{QC}=\dfrac{FB}{EC}(*)$ +Indeed,According to Menelaus's Theorem: - $\dfrac{LC}{LB}.\dfrac{FB}{FA}.\dfrac{EA}{EC}=1 \Rightarrow \dfrac{FB}{EC}=\dfrac{LB}{LC}(i)$ - $\dfrac{MA}{MC}.\dfrac{PB}{PA}.\dfrac{LC}{LB}=1 \Rightarrow \dfrac{NB}{NC}.\dfrac{QB}{QC}.\dfrac{LC}{LB}=1(ii)$ +According to $(i);(ii)$.We'll have $RS//BC$ +The next step,We'll prove $F;E$ is tangency point of incircle $\triangle ARS$ with $AB;AC(**)$ +$\dfrac{PF}{AF}=\dfrac{PQ}{SA}=\dfrac{PB}{AR} \Rightarrow F$ is tangency point of incircle $\triangle ARS$ with $AB$.Similarly,We'll get $(**)$ +By chasing angle we'll get the result P/S:My solution is a bad idea

let $X,Y$ be tangency points of the $ABC$ incircle with $AB,AC$ than since $BPQ$ and $BAC$ are homothetic $CX||QF$ and similarly $BY||RE$ so $AC/AS=AX/AF=AY/AE=AB/AR$ so $BC||RS$ therefore $ABC$ and $ARS$ are homothetic since $X,Y$ are tangency points of the $ABC$ incircle with $AB,AC$ and $AF/AX=AE/AY=AR/AB$ than $E,F$ are tangency points of $ARS$ incircle with $AR,AS$ if $I$ is the incentre of $AEF$ than $\angle EIF=90+\angle EAF/2=90+\angle RAS/2$ and if $H$ is the tangency point of $ARS$ incircle with $RS$ than $\angle FHE=90-\angle RAS/2$ than $\angle FHE+\angle FIE=180$ so $I$ is on the incircle of $ARS$ (sry for posting the same solution as above i didn't read the previous posts before posting mine)

Let $S'$ be a point on $AC$ such that the homothety centered at $E$ sends $MC$ to $AS'$. Notice that $$AS'=\frac{AE\cdot MC}{ME}=\frac{AF\cdot PQ}{PF}=AS,$$so $S'=S$. It follows that the homothety centered at $E$ sends $\triangle MNC$ to $\triangle ARS$. Therefore the incircle of $\triangle ARS$ is tangent to $AB$ and $AC$ at $F$ and $E$ since $AF=AE$. Let the midpoint of minor arc $FE$ of the incircle be $K$. Then $\angle ABK = \angle KEF = \angle KFE$ and the result follows. $\Box$

USA TST 2010/5 wrote: Let $ABC$ be a triangle. Point $M$ and $N$ lie on sides $AC$ and $BC$ respectively such that $MN || AB$. Points $P$ and $Q$ lie on sides $AB$ and $CB$ respectively such that $PQ || AC$. The incircle of triangle $CMN$ touches segment $AC$ at $E$. The incircle of triangle $BPQ$ touches segment $AB$ at $F$. Line $EN$ and $AB$ meet at $R$, and lines $FQ$ and $AC$ meet at $S$. Given that $AE = AF$, prove that the incenter of triangle $AEF$ lies on the incircle of triangle $ARS$. Let the incircles $\omega_b, \omega_c$ of $\triangle BPQ$ and $\triangle CMN$ touch $BC$ at $X, Y$ and $PQ, MN$ at $Z, T$ respectively. Let $\omega$ be the circle passing through $E, F$ and tangent to $AE, AF$. Note that $PQ \parallel AC$ so $Z, E$ are bottom-points in $\omega, \omega_c$ w.r.t $AC$; hence the dilation at $F$ that takes $\omega$ to $\omega_c$ takes $Z$ to $E$. Consequently, $P \mapsto A$ and $Q \mapsto S$ under this dilation. Let $V$ be the bottom-point of $\omega$ wrt $BC$; then line $VS$ touches $\omega$. Analogously, line $VR$ touches $\omega$ and $VR \parallel BC \parallel VS$ so $RS$ is tangent to $\omega$, hence it is the incircle of $\triangle ARS$. Finally, if $I$ is the incenter of $\triangle AEF$ then $$\measuredangle EIF+\measuredangle EVF=\left(90^{\circ}-\tfrac{1}{2}A\right)+\left(90^{\circ}+\tfrac{1}{2}A\right)=180^{\circ},$$just as we desired. $\blacksquare$

Let: $D' , E' ,F'$ are the tangency points in $ABC $ and let $X,Y$ the tangency points in $BC$ in the triangles $\triangle PBQ , \triangle CMN$ claim(1): $RS || BC$ from minilaus $\frac{BF}{AF}.\frac{AS}{SC}.\frac{CQ}{QB}=1=\frac{CE}{AE}.\frac{AR}{BR}.\frac{BN}{NC}$ but we have from similarity $\triangle BQP \sim \triangle BAC $ $\frac{BF.CQ}{BQ}=FF'=EE'=\frac{CE.NB}{NC} \implies \frac{AS}{SC}=\frac{AR}{RB}$ claim(2): $F,E$ are the tangency points in $\triangle ARS$ let $\tau_1$ is the homothety $(B: \triangle BPQ $ ⟶ $\triangle BCA)$ let $\tau_2$ is the homothety $(C: \triangle CBA $ ⟶ $\triangle CMN)$ let $\tau_3$ is the homothety $(F: \triangle BQP $ ⟶ $\triangle RSA)$ let $\tau_4$ is the homothety $(E: \triangle RSA $ ⟶ $\triangle CMN)$ notice $\tau_3(\tau_4(\triangle BPQ))=\triangle CMN$ and $\tau_1(\tau_2(\triangle BPQ))=\triangle CMN$ Now let $D \in RS $ such that $RD=RF$ the $\tau_3(X)=D$ but we have $\tau_1(X)=D' , \tau_2(D')=Y $ thus $\tau_4(D)=Y \implies ES=SD$ then $D,E,F$ are the tangency point in $ARS$ Finally let $I,J$ be the incenters of $\triangle ARS, \triangle AEF$ $\angle JEF=\frac{90-A}{2}=2\angle JIF \implies J \in (DEF)$ and we win

MellowMelon wrote: Let $ABC$ be a triangle. Point $M$ and $N$ lie on sides $AC$ and $BC$ respectively such that $MN || AB$. Points $P$ and $Q$ lie on sides $AB$ and $CB$ respectively such that $PQ || AC$. The incircle of triangle $CMN$ touches segment $AC$ at $E$. The incircle of triangle $BPQ$ touches segment $AB$ at $F$. Line $EN$ and $AB$ meet at $R$, and lines $FQ$ and $AC$ meet at $S$. Given that $AE = AF$, prove that the incenter of triangle $AEF$ lies on the incircle of triangle $ARS$. More of a computational proof - Let $AE=AF=x$ , now just use Menalaus and get $AR=\frac{cx}{s-a}$ and $AS=\frac{bx}{s-a}$. Let incircle of $ABC$ touch at $X,Y,Z$ So homothety with center $A$ and ratio $\frac {s-a}{x}$ takes $E,F$ to $Y,Z$ and $R,S$ to $B,C$ Now if $J$ is incenter of $AYZ$ , we have $\angle YJZ=180-2\angle JYZ=180-\angle ZYA= 180-\angle YXZ$ And so $J$ lies on incircle of $ABC$ Hence proved.

Since$$\frac{AR}{AB}=\frac{AR}{MN}\frac{MN}{AB}=\frac{EA}{EM}\frac{CM}{AC}=\frac{AE}{AC}\frac{AC}{s-BC}=\frac{AF}{s-BC}$$where $s$ is the semiperimeter of $\triangle ABC$ and similarly $\frac{AS}{AC}=\frac{AE}{s-BC}$, the incircle of $\triangle ARS$ is tangent at $E,F$ and the desired conclusion follows (say, by fact 5 on $\triangle AEF$).

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.145279304834013cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -0.461015534670835, xmax = 24.902182727414193, ymin = -9.228339021613003, ymax = 4.013554781678436; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pair A = (6.193881397138292,2.4906905793222758), B = (4.077953463373513,-3.5630078158232457), C = (14.031953463373508,-3.6530078158232455), P = (4.643066511512965,-1.9462117529674035), Q = (6.7364255178406705,-3.5870446336755704), F = (4.360310434311947,-2.7551807085990117), M = (8.920731017020406,0.3533101760926215), R = (2.5293275313476555,-7.993647522301948), I_1 = (6.9967101938426755,-0.25087995692253895), G = (8.078338025809694,-8.043819407423667), H = (1.6758389690443611,-3.5412888782614993), J = (8.230343365277912,-1.6218982195664913), K = (5.31752816097839,-2.474873510398549), I = (9.223017156441804,-0.6789631762770663), L = (9.630278634925359,-3.613209670900568); draw(A--B--C--cycle, linewidth(0.8) + ffdxqq); /* draw figures */ draw(A--B, linewidth(0.8) + ffdxqq); draw(B--C, linewidth(0.8) + ffdxqq); draw(C--A, linewidth(0.8) + ffdxqq); draw(circle((9.070294153452629,-1.9113274930727953), 5.258470638349202), linewidth(1.2) + linetype("0 3 4 3")); draw(circle((4.939214175898718,-2.9575228997406544), 0.6132470173911679), linewidth(1.2) + linetype("4 4") + red); draw((19.768557940853196,-8.149517598337273)--F, linewidth(0.4)); draw(C--(19.768557940853196,-8.149517598337273), linewidth(0.4)); draw(circle((9.643816061233105,-2.115944365536324), 1.4973265029907363), linewidth(1.2) + linetype("4 4") + red); draw(B--R, linewidth(0.4)); draw((10.56752142016272,-0.9374898409817369)--R, linewidth(0.4)); draw(R--(19.768557940853196,-8.149517598337273), linewidth(0.4)); draw(circle((8.114291962675042,-4.06729668468894), 3.9766852591073545), linewidth(1.2) + linetype("4 4") + ffxfqq); draw((10.56752142016272,-0.9374898409817369)--F, linewidth(0.4)); draw(Q--P, linewidth(0.4)); draw((7.540930186084854,-3.594318636100925)--M, linewidth(0.4)); draw(F--I_1, linewidth(0.4)); draw((10.56752142016272,-0.9374898409817369)--I_1, linewidth(0.4)); draw(H--F, linewidth(0.4)); draw(B--H, linewidth(0.4)); draw(J--I, linewidth(0.4)); draw(H--M, linewidth(0.4)); draw(G--(10.56752142016272,-0.9374898409817369), linewidth(0.4)); /* dots and labels */ dot(A,blue); label("$A$", (6.262520532316235,2.6562212094697784), NE * labelscalefactor,blue); dot(B,blue); label("$B$", (3.6741169760113537,-3.814787681292427), NE * labelscalefactor,blue); dot(C,blue); label("$C$", (14.390739017054125,-3.546477556553506), NE * labelscalefactor,blue); dot(P,blue); label("$P$", (4.49483029874217,-1.7314384774372777), NE * labelscalefactor,blue); dot(Q,linewidth(4.pt) + blue); label("$Q$", (6.120473995689748,-4.035748960489185), NE * labelscalefactor,blue); dot(F,linewidth(4.pt) + blue); label("$F$", (4.037124791834599,-2.7415471823367437), NE * labelscalefactor,blue); dot((19.768557940853196,-8.149517598337273),linewidth(4.pt) + blue); label("$S$", (19.693809717776322,-8.707501720649216), NE * labelscalefactor,blue); dot((10.56752142016272,-0.9374898409817369),linewidth(4.pt) + blue); label("$E$", (10.634397270709236,-0.8160274636221365), NE * labelscalefactor,blue); dot(M,linewidth(4.pt) + blue); label("$M$", (8.97718767673355,0.47817431453030446), NE * labelscalefactor,blue); dot((7.540930186084854,-3.594318636100925),linewidth(4.pt) + blue); label("$N$", (7.304195134243809,-3.1834697407302603), NE * labelscalefactor,blue); dot(R,linewidth(4.pt) + blue); label("$R$", (2.3641322493448587,-8.53388928699462), NE * labelscalefactor,blue); dot(I_1,linewidth(4.pt) + blue); label("$I_1$", (6.625528348139481,-0.011097089405374476), NE * labelscalefactor,blue); dot(G,linewidth(4.pt) + blue); label("$G$", (8.14069140548868,-8.549672235508675), NE * labelscalefactor,blue); dot(H,linewidth(4.pt) + blue); label("$H$", (1.385589441473501,-3.2939503803286394), NE * labelscalefactor,blue); dot(J,linewidth(4.pt) + blue); label("$J$", (7.825032435207596,-1.5104771982405196), NE * labelscalefactor,blue); dot(K,linewidth(4.pt) + blue); label("$K$", (5.457590158099473,-2.15757808731674), NE * labelscalefactor,blue); dot(I,linewidth(4.pt) + blue); label("$I$", (9.292846647014633,-0.5477173388832158), NE * labelscalefactor,blue); dot(L,linewidth(4.pt) + blue); label("$L$", (9.81368394797842,-3.9884001149470225), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] I have a different solution than others. first note that since $AE=AF$ then $E$ and $F$ are $\textit{anti-homologous}$ points in red circles, Furthermore $IJ||I_1P$. now if we prove that $BC||RS$ then by homothety with center $E$ we are done. which is trivial by $\textit{desaurgue's theorem}$ on $\triangle CEN$ and $\triangle BKQ$. now it is trivial that $EIJL \sim EI_1FG$ hence we are done.$\blacksquare$

We first contend that $BC\parallel RS$. By Menelaus(twice), we have \begin{align*} \frac{AF}{FB}\cdot\frac{BQ}{QC}\cdot\frac{CS}{SA}=1,\qquad \frac{AAE}{EC}\cdot\frac{CN}{NB}\cdot\frac{BR}{RA}=1. \end{align*}Since $AF=AE$, this means \begin{align*} \frac{FB\cdot QC\cdot SA\cdot CN\cdot BR}{BQ\cdot CS\cdot EC\cdot NB\cdot RA}=1\implies\frac{FB\cdot QC\cdot CN}{BQ\cdot EC\cdot NB}=\frac{CS\cdot RA}{SA\cdot BR}. \end{align*}We will prove the LHS is equal to $1$; evidently, this implies our claim. We have \begin{align*} \frac{FB}{BQ}\cdot\frac{CN}{EC}=\frac{s-b}{s}\cdot\frac{s}{s-c}=\frac{s-b}{s-c}. \end{align*}Hence, it suffices to prove that $\tfrac{QC}{NB}=\tfrac{s-c}{s-b}$. Let us first define $BQ=ka, CN=\ell a$ for constants $k, \ell\leq 1$. Because $AF=AE$, we have \begin{align*} c-(s-b)k=b-(s-c)\ell\implies k=\frac{c-b+(s-c)\ell}{s-b}. \end{align*}Therefore, \begin{align*} \frac{QC}{NB}&=\frac{(1-k)a}{(1-\ell)a} \\ &=\frac{\frac{s-b}{s-b}-\frac{c-b+(s-c)\ell}{s-b}}{1-\ell} \\ &=\frac{(s-c)(1-\ell)}{(s-b)(1-\ell)} \\ &=\frac{s-b}{s-c}, \end{align*}as desired. Now, notice that homothety at $F$ and $E$ implies that the incircle of $\triangle ARS$ touches sides $AR$ and $AS$ at $F$ and $E$ respectively. This allows us to reduce a big chunk of the points and reformulate the problem as such: USA TST 2010, Reformulated wrote: Let $\triangle ABC$ have intouch points $D, E, F$ and incenter $I$. Denote the incenter of $\triangle AEF$ as $J$. Prove $J$ lies on the incircle of $\triangle ABC$. We redefine $J$ to be $AI\cap FE=J$. But now, it is clear that $J$ is the incenter of $\triangle AEF$ by the (Converse of the) Incenter Excenter Lemma. edit: this is wrong, i'm fixing it rn edit2: this should be right now.

Quite nice. Denote by $\omega$ the circle tangent to $\overline{AR}$ and $\overline{AS}$ at $F$ and $E$ respectively. I will show that $\omega$ is the incircle of triangle $ARS$, from where the problem becomes a straightforward angle chase. Let $X$ and $Y$ be the touchpoints on $\overline{BQ}$ and $\overline{CR}$. Let $\omega_1$ and $\omega_2$ be the incircles of $BPQ$ and $CMN$. Obviously $\omega_1$ and $\omega$ are tangent, and so are $\omega_2$ and $\omega$. Thus $\overline{FX}$ and $\overline{EY}$ concur at a point $D$ on $\omega$. Now construct the line $\ell$ through $D$ parallel to $\overline{BC}$. As $D$ is the arc midpoint of the arc on $\omega$ cut by $\overline{BC}$, $\ell$ is a tangent to $\omega$. So it suffices to show that $R$ and $S$ lie on $\ell$. To do this, consider the homothety at $F$ that takes $\omega_1$ to $\omega$. Because $\frac{FP}{FA} = \frac{FQ}{FS}$, it also takes $Q$ to $S$, thus $S$ lies on $\ell$. Similarly, $R$ lies on $\ell$, so we're done.

Huh Let $E'$ and $F'$ be the intouch points of $\triangle ABC$ on $AC$ and $AB$. Then $\triangle FAS\sim\triangle FPQ\sim\triangle F'AC$ and $\triangle EAR\sim\triangle EMN\sim\triangle E'AB$, so degenerate pentagons $AF'BCE$ and $AFRSE$ are homothetic at $A$. Hence, $E$ and $F$ are the intouch points of $\triangle ARS$, so if $I$ is the incenter of $\triangle AEF$ and $D$ is the intouch point on $RS$ we have $$\angle EIF = 90^\circ + \frac{1}{2}\angle A = 180^\circ - \angle EDF$$ as desired.