First Solution (homothety): Let $Z$ be the diametrically opposite point on the incircle. We claim this is the desired intersection. [asy][asy]size(8cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); draw(A--B--C--cycle,red); draw(incircle(A, B, C), lightblue); pair D = 0.35*B+0.65*C; draw(A--D, red); pair I_B = incenter(A, B, D); pair I_C = incenter(A, C, D); pair E = foot(I_B, B, C); pair F = foot(I_C, B, C); draw(incircle(A, B, D), heavygreen); draw(incircle(A, C, D), heavygreen); pair I = incenter(A, B, C); pair P = extension(I_B, I_C, A, D); draw(I_B--I_C, heavygreen); pair X = extension(B, I, C, P); pair Y = extension(C, I, B, P); pair Z = extension(E, X, F, Y); draw(B--I--C, red); draw(X--C, dotted+heavygreen); draw(Y--B, dotted+heavygreen); draw(E--Z--F, dashed+heavygreen); pair T = extension(B, C, I_B, I_C); draw(I_C--T, dotted+heavygreen); draw(C--T, dotted+red); pair W = foot(I, B, C); dot("$A$", A, dir(A)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$D$", D, dir(D)); dot("$I_B$", I_B, dir(170)); dot("$I_C$", I_C, dir(30)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$I$", I, dir(90)); dot("$P$", P, dir(250)); dot("$X$", X, dir(160)); dot("$Y$", Y, dir(20)); dot("$Z$", Z, dir(140)); dot("$T$", T, dir(T)); dot("$W$", W, dir(W)); [/asy][/asy] Note that: $P$ is the insimilicenter of $\omega_B$ and $\omega_C$ $C$ is the exsimilicenter of $\omega$ and $\omega_C$. Thus by Monge theorem, the insimilicenter of $\omega_B$ and $\omega$ lies on line $CP$. This insimilicenter should also lie on the line joining the centers of $\omega$ and $\omega_B$, which is $\overline{BI}$, hence it coincides with the point $X$. So $X \in \overline{EZ}$ as desired. Second Solution (harmonic): Let $T = \overline{I_B I_C} \cap \overline{BC}$, and $W$ the foot from $I$ to $\overline{BC}$. Define $Z = \overline{FY} \cap \overline{IW}$. Because $\angle I_B D I_C = 90^{\circ}$, we have \[ -1 = (I_B I_C; PT) \overset{B}{=} (I I_C; YC) \overset{F}{=} (I\infty; ZW) \]So $I$ is the midpoint of $\overline{ZW}$ as desired. Third Solution (outline, barycentric, Andrew Gu): Let $AD = t$, $BD = x$, $CD = y$ (so with Stewart). We then have $D = (0:y:x)$ and so \[ \overline{AI_B} \cap \overline{BC} = \left( 0 : y + \frac{tx}{c+t} : \frac{cx}{c+t} \right) \]hence intersection with $BI$ gives $I_B = (ax : cy+at : cx)$. Similarly, $I_C = (ay : by : bx+at)$. Then, we can compute \[ P = \left( 2axy : y(at+bx+cy) : x(at+bx+cy) \right) \]since $P \in \overline{I_B I_C}$, and clearly $P \in \overline{AD}$. Intersection now gives \begin{align*} X &= \left( 2ax : at+bx+cy : 2cx \right) \\ Y &= \left( 2ay : 2by : at+bx+cy \right). \end{align*}From here one can check that the antipode \[ Q = \left( 4a^2 : -a^2+2ab-b^2+c^2 : -a^2+2ac-c^2+b^2 \right) \]lies on each of lines $EX$ and $FY$ (using Stewart's Theorem).

My solution ( Basically the same as v_Enhance's solution ) : Observe that $P$ is the insimilicenter of $\omega_B$ and $\omega_C$, and $C$ is the exsimilicenter of $\omega_C$ and $\omega$. So the line joining them contains the insimilicenter of $\omega$ and $\omega_B$, which must be on $BI$. So $X$ is the insimilicenter of the mentioned circles. By the negative homothety at $X$, tangents to the circles at sent to parallel tangents to the other circle, so noticing that $BC$ is sent to the tangent to the incircle parallel to it, $E$ is sent to the antipode of the incircle touchpoint with $BC$ with respect to $\omega$.

Let $I_B$ and $I_C$ the centers of $\omega_B$ and $\omega_C$, let $r_b$ and $r_c$ the radius of $\omega_B$ and $\omega_C$, let $r$ the inradius of $\triangle ABC$ and $G$ the foot of the perpendicular of $I$ in $BC$ and $H$ is the reflection of $G$ in $I$. By the Menalaus' theorem in $\triangle II_BI_C$ wrt $\overline {XPC}$ we get: $$1=\frac{IX}{XI_B}\cdot \frac{I_BP}{PI_C}\cdot \frac{CI_C}{CI}=\frac{IX}{XI_B}\cdot\frac{r_b}{r_c}\cdot \frac{r_c}{r} \Longrightarrow \frac{IX}{XI_B}=\frac{r}{r_b}$$$\Longrightarrow$ $\frac{IH}{I_BE}=\frac{r}{r_b}=\frac{IX}{XI_B}$, so from $IH\parallel I_BE$ we get $E$, $X$ and $H$ are collinear, similarly $F$, $Y$ and $H$ are collinear, hence $EX$ and $FY$ meet on the incircle of $\triangle ABC$.

Let $O_B,O_C$ be the centers of $\omega_B$ and $\omega_C$, $\{T\}=EX\cap FY,\ \{Q\}=O_B O_C\cap BC,\ \{R\}=IP\cap BC$ and $J$ the tangency point of $(I)$ with $BC$. $P,Q$ are the centers of similitude between $\omega_B$ and $\omega_C$ so $\dfrac{PO_B}{PO_C}=\dfrac{O_B E}{O_C F}=\dfrac{QO_B}{QO_C}$ or equivalently $(Q,P,O_B,O_C)=-1$. We have that $I(Q,P,O_B,O_C)=-1 \Leftrightarrow (Q,R,B,C)=-1$, so $Q-X-Y$ are collinear. Therefore $\triangle{O_B XE}$ and $\triangle{O_CYF}$ are perspective so by Desargues we have $TI\perp BC$, implying $T-I-J$ collinear. Finally, $P(Q,R,B,C)=-1\Leftrightarrow (O_B,I,B,X)=-1\Leftrightarrow E(O_B,I,B,X)=-1\Leftrightarrow (EO_B,EI,EJ,ET)=-1$, implying that $I$ is the midpoint of $JT$ and thereby $T\in (I)$.

This problem could be derived from problem 4 of Romanian Masters in Mathematics 2015 https://artofproblemsolving.com/community/c6h1058212p4575366

I heard you can affine transform to a right isosceles triangle and use mass points, can someone confirm?

I don't see how affine transforms don't preserve circles?

EulerMacaroni wrote: I heard you can affine transform to a right isosceles triangle and use mass points, can someone confirm? care to explain the inside joke?

All you care about is ratios inside triangle BIC (specifically IY:IC) which can be done with mass points. Or, affine transform BIC to a right isosceles triangle and use coordinates to easily calculate ratios. (Circles in this problem are pretty irrelevant, only centers and ratios are relevant)

linpaws wrote: All you care about is ratios inside triangle BIC (specifically IY:IC) which can be done with mass points. Or, affine transform BIC to a right isosceles triangle and use coordinates to easily calculate ratios. (Circles in this problem are pretty irrelevant, only centers and ratios are relevant) could u be more explicit im not sure I get it

Easy for a TSTST/5 CantonMathGuy wrote: Let $ABC$ be a triangle with incenter $I$. Let $D$ be a point on side $BC$ and let $\omega_B$ and $\omega_C$ be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Suppose that $\omega_B$ and $\omega_C$ are tangent to segment $BC$ at points $E$ and $F$, respectively. Let $P$ be the intersection of segment $AD$ with the line joining the centers of $\omega_B$ and $\omega_C$. Let $X$ be the intersection point of lines $BI$ and $CP$ and let $Y$ be the intersection point of lines $CI$ and $BP$. Prove that lines $EX$ and $FY$ meet on the incircle of $\triangle ABC$. Proposed by Ray Li Let $D_1$ be the tangency point of $\omega = \text{incircle}$ of $\triangle ABC$ with side $BC$ and $D_1^{\prime}$ be its antipode. Also let $T = \overline{XE} \cap \overline{YF}$. Claim : $D_1 - I - T$ Proof. Let $O_bO_c \cap BC = Q \implies Q$ is the ex-similicenter of $\{\omega_B , \omega_C\}$ by Monge's theorem. Next, notice that $X$ lies on the line joining $C$ and $P$ which are respectively the ex-similicenter $\{\omega_C , \omega\}$ and in-similicenter $\{\omega_B , \omega_C\}$. So it follows that $X$ is the in-similicenter of $\{\omega , \omega_B\}$. Similarly $Y$ is the in-similicenter of $\{\omega , \omega_C\}$. Again by Monge's theorem it follows that $X - Y - Q$. Therefore $\triangle XO_BE$ and $\triangle YO_CF$ are perspective $\implies D_1-I-T$. Now observe that $$-1 = (BX ; O_BI) \overset{E}{=} (D_1T ; P_{\infty}I) \iff T \equiv D_{1}^{\prime} \blacksquare$$

CantonMathGuy wrote: Let $ABC$ be a triangle with incenter $I$. Let $D$ be a point on side $BC$ and let $\omega_B$ and $\omega_C$ be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Suppose that $\omega_B$ and $\omega_C$ are tangent to segment $BC$ at points $E$ and $F$, respectively. Let $P$ be the intersection of segment $AD$ with the line joining the centers of $\omega_B$ and $\omega_C$. Let $X$ be the intersection point of lines $BI$ and $CP$ and let $Y$ be the intersection point of lines $CI$ and $BP$. Prove that lines $EX$ and $FY$ meet on the incircle of $\triangle ABC$. Proposed by Ray Li Assume wlog that $\angle BAD>\angle CAD.$ Let $I_1, I_2$ be the centers of $\omega_B, \omega_C$ respectively. Then if we define $R=I_1I_2 \cap BC,$ then since $\angle I_1DI_2=\pi/2$ and $\angle PDI_2=\angle I_2DR,$ hence $(I_1I_2,PR)$ is harmonic. 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IMO it was really easy for a P5. USTSTST 2017 P5 wrote: Let $ABC$ be a triangle with incenter $I$. Let $D$ be a point on side $BC$ and let $\omega_B$ and $\omega_C$ be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Suppose that $\omega_B$ and $\omega_C$ are tangent to segment $BC$ at points $E$ and $F$, respectively. Let $P$ be the intersection of segment $AD$ with the line joining the centers of $\omega_B$ and $\omega_C$. Let $X$ be the intersection point of lines $BI$ and $CP$ and let $Y$ be the intersection point of lines $CI$ and $BP$. Prove that lines $EX$ and $FY$ meet on the incircle of $\triangle ABC$. Proposed by Ray Li Let $I_BI_C\cap BC=T$. Note that $\{T,P\}$ are the exsimillicenter and insimillicenter of $\{\omega_B,\omega_C\}$ respectively. Hence, $(TP;I_BI_C)$ is harmonic. So, $(TP;I_BI_C)\overset{X}{=}(YC;I_CI)\overset{T}{=}(XB;I_BI)$.So, by the so-called Prism Lemma we get that $\{XY,I_BI_C,BC\}$ concurs at a point $T$. Let $M\in BC$ such that $IM\perp BC$ and let $IM\cap YF=K'$. So, $(YC;II_C)\overset{F}{=}(K'M;I'\infty_{MK'})\implies K\in\odot(I)$. Let $XE\cap IM=K'''$. So, $(BX;II_B)\overset{E}{=}(MK''';I\infty_{MK'''})\implies K'''\in\odot(I)$. So, $K'\equiv K'''\equiv K\in\odot(I).$ $\blacksquare$

Sorry for bumping an old thread, but what are affine transformations, insimilicenter and exsimilicenter?

The insimilicenter and exsimilicenter between two circles are the intersection of the internal common tangents and the intersection of the external common tangents, respectively. The insimilicenter is the center of the negative homothety between the two circles, and the exsimilicenter is the center of the positive homothety. Not sure what an affine transformation is but it seems to have something to do with projective geometry. However, projective isn't really needed for this problem, as Monge's theorem pretty much kills it.

An affine transformation (on the real projective plane) is a bijection that preserves collinearity and ratios (of parallel segments). (It is a projective transformation that preserves the line at infinity.) In general there exists a unique affine transformation sending any triangle to any other triangle.

eaSy fOR p5 Let $I_1$ and $I_2$ be the centres of $\omega_B$ and $\omega_C$ respectively. Define $Q=I_1I_2\cap BC$, and let $R$ be the perpendicular foot from $I$ to $BC$, and let $Z=EX\cap IR$. (We assume $Q,B,C$ lie in that order; the other case is similar.) Then $$-1=(Q,P;I_1,I_2)\stackrel{C}{=}(B,X;I_1,I)\stackrel{E}{=}(R,Z;I_\infty,I).$$(the first equality holds since $I_1D\perp I_2D$ and $I_1D$ bisects $\angle PDQ$). If $Z'=FY\cap IR$, we similarly have $$-1=(Q,P;I_1,I_2)\stackrel{B}{=}(C,Y;I_2,I)\stackrel{F}{=}(R,Z';I_\infty,I).$$It follows that $Z=Z'$ is the point diametrically opposite $R$ on the incircle. Hence $EX$ and $FY$ concur on the incircle as desired.

Assume that $J$ is the exsimilicenter of $\omega_B$ and $\omega_C$. We will to show that $-1=(O_BO_C;PJ)$. Note that $DO_B$ and $DO_C$ bisect angles $\angle PDJ$ internally and externally respectively. Hence we have $-1=(O_BO_C;PJ)$. This means that $P$ is the insimilicenter of $\{\omega_B,\omega_C\}$. Denote by $\omega$ the incircle of $'\triangle ABC$ and by $R$ the antipode of the point of tangency of $\omega$ with $BC$ wrt $\omega$. Cross ratio chase gives : $$-1=(O_bO_C;PJ) \overset{B}{=} (IO_C;CY)$$Note that since $C$ is the exsimilicenter of $\omega$ and $\omega_C$ , hence $Y$ is the insimilicenter of $\omega$ and $\omega_C$. Similarly , $X$ is the insimilicenter of $\omega$ and $\omega_B$. We are ready to finish. Note that the negative homothety at $X$ carrying $\omega_B\mapsto \omega$ sends $E\mapsto R$ so $X\in ER$. Similarly $Y \in RF$ and we are done. $\blacksquare$

Let $\omega_Z$ be the incircle, and if $\omega_Z$ is tangent to $BC$ at $S$, let $Z$ be the antipode of $S$. Let $X'$ be the center of negative homothety from $\omega_B$ to $\omega_Z$. Consider the maps \begin{align} h_{12}\colon \omega_B \to \omega_Z \text{ centered at } X'\\ h_{23}\colon \omega_Z\to \omega_C \text{ centered at } C\\ h_{13}\colon \omega_B \to \omega_C \text{ centered at } P \end{align}Thus, $X'$ lies on $PC$, and $X'$ also lies on $BI$ the line between the centers of the circle, so $X'=PC\cap BI=X$. To finish, we claim that $EX,FY$ concur at $Z$. To see this, note that in homothety $h_{12}$ $E$ is sent to $Z$, so $X$ lies on $EZ$ and we're done. $\blacksquare$.

it can be easily proved by Monge Theroem

Let $\omega$ be the incircle, $A_1$ be the $BC$ contact point of the incircle, $T=I_1I_2 \cap BC$ and $Z = \omega \cap EX$. Claim: $Z$ is the antipode of $A_1$. Proof: $$-1=(I_1I_2;PT) \stackrel{B}= (II_2;YC) \stackrel{F}= (IP_{\infty};ZA_1)$$$\blacksquare$ Symmetrically $FY \cap \omega$ also is the antipode so they indeed are concurrent on the incircle. Here is a super accurate diagram...

Let $M$ be the point where the incircle touches $BC$, and $N$ be the $M-$ antipode w.r.t the incircle.We Claim that the required concurrency point on the incircle is precisely $N$.Let $\omega_B$ and $\omega_C$ denote the incircles. Let $r_1$ and $r_2$ be the radius of the incircles of $ABD$ and $ACD$ , respectively.Then since $P$ lies on $I_BI_C$ and also the internal common tangent , $$\frac{I_BP}{I_CP}= \frac{r_1}{r_2}$$Let $I_BI_C \cap BC =L$.Then similarity gives that $$\frac{LI_B}{LI_C}= \frac{r_1}{r_2} \implies -1= (L,P;I_B,I_C) \stackrel{C}=(B,X;I_B,I)$$ Now , if $X$ were the intersection of $I_BI$ and $EN$ (which is what we claimed), then $X$ would also lie on the line through $M$ and the $E-$ antipode in $\omega_B$.Then , similarity ratios would give that $$\frac{I_BX}{IX}=\frac{r_1}{r}$$showing which alone would be sufficient to prove our claim. But $$-1 =(B,X;I_B,I) \implies \frac{I_BX}{IX}=\frac{BI_B}{BI} =\frac{r_1}{r}$$as desierd . Thus $EX$ and $FY$ intersect on the $M-$ antipode w.r.t the incircle of $\triangle ABC$.

Let $I,I_B,I_C,$ be the centers of $\omega,\omega_B,\omega_C,$ let $Z=\omega\cap\overline{BC},$ and let $Z'$ be the antipode of $Z.$ Notice $P$ is the insimilicenter of $\omega_B$ and $\omega_C.$ Claim: $X$ is the insimilicenter of $\omega$ and $\omega_B.$ Proof. By Monge on $\omega_B,\omega_C,\omega,$ taking the insimilicenters of $(\omega,\omega_C),(\omega_B,\omega_C),$ and the exsimilicenter of $(\omega,\omega_B),$ we know the insimilicenter of $(\omega,\omega_B)$ lies on $\overline{CP}.$ It suffices that $X=\overline{CP}\cap\overline{II_B}.$ $\blacksquare$ Similarly, $Y$ is the insimilicenter of $(\omega,\omega_C).$ Consider the negative homothety $\varphi$ at $X$ such that $\varphi(\omega_B)=\omega.$ Note $\varphi(E)=Z'$ so $Z'$ lies on $\overline{EX}.$ Similarly, $Z'$ lies on $\overline{FY}.$ $\square$

Anyways, really nice problem. $I_1,I_2$ be the centers of $\omega_B$ and $\omega_C$. $\odot(I)$ denote the incircle. Also let $T=I_1I_2\cap BC$. Now as $AD$ is a common internal tangent to both circles, so $P=I_1I_2\cap AD$ becomes the insimilicenter of $\{\omega_B,\omega_C\}$. Also note that $B$ and $C$ are the exsimilicenters of $\{\omega_B,\odot(I)\}$ and $\{\omega_C,\odot(I)\}$. Easy to notice that $\overline{B-I_1-I}$ and $\overline{C-I_2-I}$ are collinear. So Monge-D-Alembert gives $\overline{\text{in}(\omega_B,\odot(I))-\text{in}(\omega_B,\omega_C)-\text{ex}(\omega_C,\odot(I))}$ are collinear $\implies\overline{\text{in}(\omega_B,\odot(I))-P-C}$ are collinear and from the fact $\overline{\text{in}(\omega_B,\odot(I))-B-I}$ are collinear $\implies \text{in}(\omega_B,\omega_C)=PC\cap BI=X$. Similarly, $Y\equiv \text{in}(\omega_C,\odot(I))$. Also, by Monge-D-Alembert, we have $\overline{\text{in}(\omega_B,\odot(I))-\text{in}(\omega_C,\odot(I))-\text{ex}(\omega_B,\omega_C)}$ are collinear but as $\text{ex}(\omega_B,\omega_C)=I_1I_2\cap BC\implies T=BC\cap I_1I_2\cap XY$. Now $\{r_1,r_2,r\}$ be the radiuses of $\{\omega_B,\omega_C,\odot(I)\}$. Now we construct two homotheties $\Phi$ and $\Psi$. $\Phi$ has center $X$ and scale $\dfrac{-r}{r_1}$ and $\Psi$ has center $Y$ and scale $\dfrac{-r_2}{r}$. Now note that $\omega_B\xmapsto{\Phi}\odot(I)$ and $\odot(I)\xmapsto{\Psi}\omega_C$. So center of $\Phi\circ\Psi\in I_1I_2$ and has scale $\dfrac{-r}{r_1}\cdot\dfrac{-r_2}{r}=\dfrac{r_2}{r_1}$. Now see that $\omega_B\xmapsto{\Phi}\odot(I)\xmapsto{\Psi}\omega_C$. Thus the center of $\Phi\circ\Psi$ is either the insimilicenter or the exsimilicenter of $\{\omega_B,\omega_C\}$ and as the scale is positive, we conclude that it is the exsimilicenter. Thus $\Phi\circ\Psi$ has center $T$ and maps $\omega_B\xmapsto{\Phi\circ\Psi}\omega_C$, also $BC\xmapsto{\Phi\circ\Psi}BC\implies E\xmapsto{\Phi\circ\Psi}F$. Now as $\Phi(E)\in \odot(I)$ and $\Psi^{-1}(F)\in \odot(I)$ to finish, notice we have $\Psi(\Phi(E))=F\implies \Phi(E)=\Psi^{-1}(F)$ and thus they concur at some point on the incircle.

Monge (two in, one out) gets that $X$ is the insimilicentre of $\omega_B$ and $\omega$, the incircle of $\triangle ABC$. Thus, a dilation centred at $X$ swapping $\omega_B$ and $\omega$ swaps $E$ with the 'top' point of $\omega$, say $T$ (because tangent lines to $E$ and $T$ are parallel). Thus, $E, X, T$ are collinear and similarly $F,Y,T$ are collinear.

IAmTheHazard wrote: Only about half of this solution is mine, since I got the idea for homothety from the first part of a walkthrough (it's not too difficult to motivate, but knowing myself I would've taken at least an hour to get there). This solution is mostly written to store whatever I found out about compositions of homotheties. Let $Z$ denote the antipode of the tangency point from the incircle $\omega$ to $\overline{BC}$. I claim that $Z$ is the desired intersection point. Clearly, the positive homothety sending $\omega$ to $\omega_B$ is centered at $B$. Further, by dropping perpendiculars from the centers $I_B,I_C$ (resp.) of $\omega_B,\omega_C$ to $\overline{AD}$ and using similar triangles, we find that $\tfrac{PI_B}{PI_C}=\tfrac{I_BE}{I_CF}$, so $P$ is the center of the negative homothety sending $\omega_B$ to $\omega_C$. Now consider the negative homothety sending $\omega$ to $\omega_C$; I claim that the center of this homothety is $Y$. It is known that the composition of two homotheties is a homothety (which can be proved easily with complex), so this homothety is the composition of the positive homothety sending $\omega$ to $\omega_B$ and the negative homothety sending $\omega_B$ to $\omega_C$. Given this, I will now prove the following claim about compositions of homotheties. Claim: Given a homothety $\mathcal{H}_1$ centered at $A$ sending some point $X$ to $Y$, and a homothety $\mathcal{H}_2$ centered at $B$ sending $Y$ to $Z$, the homothety $\mathcal{H}_1 \circ \mathcal{H}_2$ is centered at $\overline{AB} \cap \overline{XZ}$. Proof: Note that $\mathcal{H}_1$ sends $A$ to $A$, then $\mathcal{H}_2$ sends $A$ to some point on $\overline{AB}$, so the center of $\overline{H}_1 \circ \overline{H}_2$ lies on $\overline{AB}$. Since this center also obviously lies on $\overline{XZ}$ we're done. $\blacksquare$ Given this claim, it follows that the center of the homothety sending $\omega$ to $\omega_C$ is the intersection of $\overline{BP}$ and $\overline{II_C}$, which is just $Y$. The negative homothety sending $\omega$ to $\omega_C$ sends $Z$ to $F$, hence $F,Y,Z$ are collinear. Likewise, $E,X,Z$ are collinear, so $\overline{EX} \cap \overline{FY}=Z$ as desired. $\blacksquare$ This is the fourth problem in 3 days I have done twice, without remembering that I did it the first time. Thankfully it turns out I actually learned something in blue class geo this year so I solved it entirely by myself Let the incircle be $\omega$. By Monge on $\omega$, $\omega_B$, and the "negative radius" $\omega_C$, we find that the insimilicenter of $\omega$ and $\omega_B$ lies on $\overline{CP}$. Since it also lies on $\overline{BI}$, which joins the incenter to the center of $\omega_B$, it follows that the insimilicenter is actually $X$. Likewise, $Y$ is the insimilicenter between $\omega$ and $\omega_C$. Therefore if $Z$ is the point on $\omega$ such that the tangent to $\omega$ at $Z$ is parallel to $\overline{BC}$, then a homothety at $X$ sending $\omega_B$ to $\omega$ sends $E$ to $Z$, hence $E,X,Z$ collinear. Likewise, $F,Y,Z$ is collinear, so $\overline{EX}$ and $\overline{FY}$ meet at $Z \in \omega$. $\blacksquare$ Remark: Motivation for considering homothety was something like "well how are we going to find $\overline{EX} \cap \overline{FY}$ and prove it lies on the incircle without using homothety". Then I noticed that $P$ was the insimilicenter of $\omega_B$ and $\omega_C$ since tangency things so now we found quite a few insimilicenters and then we find more with Monge

This one's cool, it has a pretty short solution. It's evident that C and P are the respective trivial homothety centers between incircle with w_c (ex), and w_c to w_b (in); in particular, the homothety from w_b to incircle must lie on BI by definition and CP by Monge's, which is X; since X is the insimilicenter, E goes to J, where J=incircle with FY, as desired. $\blacksquare$

Consider Monge's on the incircle, $\omega_B$, and $\omega_C$. Exsimilicenter of $\omega_C$ and the incircle is $C$, insimilicenter of $\omega_B$ and $\omega_C$ is $P$. Insimilicenter of $\omega_B$ and the incircle should lie $BI$, but also $CP$ by Monge's, hence it is the point $X$. As a result, homothety at $X$ will send the bottom point of $\omega_B$, or $E$, to the top point of the incircle. Thus $EX$ will pass through this top point, as does $FY$ by symmetry. $\blacksquare$

Let $\omega$ be the incircle. Orient $\overline{BC}$ horizontally and denote the north pole and south pole of any circle as the highest and lowest points respectively. Note that $P$ is the insimilicenter of $\omega_b$ and $\omega_c$. Then Monge on $\omega_b$, $\omega_c$ and $\omega$ tells us that the insimilicenter of $\omega_b$ and $\omega$ lies on $\overline{CP}$. Similarly, the insimilicenter of $\omega_c$ and $\omega$ lies on $\overline{BP}$. However then it follows that the aforementioned insimilicenters are exactly $X$ and $Y$. However then $E$ and $F$ both map to the north pole of $\omega$ under the homotheties at $X$ and $Y$ respectively hence we are done.

Let $I_B,I_C$ be the incenters of $\triangle ABD, \triangle ACD$, respectively. Let $D_0$ be the in-touch point of side $\overline{BC}$. Let $Z=EX\cap FY$. Claim: $XY,I_BI_C,BC$ intersect at one point $J$. Proof. Let $S_B,S_C$ be the intersections of $AI_B,AI_C$ with line $BC$, respectively. Define $J$ to be the intersection of $I_BI_C$ and $BC$. Note that by internal angle bisector theorem, $\frac{S_BD}{BD}=\frac{AD}{AB+AD}$ and $\frac{DS_C}{DC}=\frac{AD}{AC+AD}$. So, $\frac{S_BD}{DS_C}=\frac{BD}{DC}\cdot\frac{AC+AD}{AB+AD}$. However, a corollary of internal angle bisector theorem tells us $\frac{AI_B}{I_BS_B}=\frac{AB+AD}{BD}$ and $\frac{AI_C}{I_CS_C}=\frac{AC+AD}{DC}$, and Menelaus' Theorem gives \[\frac{S_BJ}{S_CJ}=\frac{AI_C}{I_CS_C}\cdot\frac{I_BS_B}{AI_B}=\frac{BD}{DC}\cdot\frac{AC+AD}{AB+AD}=-\frac{S_BD}{S_CD}\]and hence $(S_BS_C;JD)=-1$. Taking perspective at $A$, we get $(I_BI_C;PJ)=-1$. However, since $P=BY\cap CX\cap I_BI_C$, we have that $I_BI_C,XY,BC$ are concurrent at $J$. $\blacksquare$ Claim: $Z,I,D_0$ are collinear. Proof. We use a lemma: For any triangle $abc$, if there are fixed points $d,e,f,g$ on sides $ab,ac,bc,bc$ and variable points $p,q$ on $ab,ac$ such that $de,pq,bc$ concur (at a fixed point), then locus of $fp\cap gq$ is the line joining $a$ and $fd\cap ge$. From this lemma, we have that $EI_B\cap FI_C,I,Z$ are collinear. However, we know that $FI_C\cap EI_B=\infty_{ID_0}$, which means $I,D_0,Z$ are collinear. $\blacksquare$ Claim: $Z$ is the reflection of $D_0$ in $I$ (or incircle), and hence lies on the incircle. Proof. We showed that $(I_BI_C;PJ)=-1$, taking perspective at $Y$ and projecting onto line $BI$, we get $(BXI_BI)=-1$. Taking perspective at $E$ and projecting onto line $ID_0$, we get $(ZD_0;I\infty_{ID_0})=-1$, which means $I$ is the mid-point of $\overline{ZD_0}$, as claimed. The proof is complete. $\blacksquare$

Note that since $P$ lies on an common internal tangent of $\omega_B$ and $\omega_C$ as well as the line joining their centers, it is the center of negative homothety betwen $\omega_B$ and $\omega_C$. Furthermore, clearly $C$ is the center of positive homothety between $\omega$ and $\omega_C$. Thus, by Monge, the center of negative homothety of $\omega_B$ and $\omega$ lies on $CP$. However, it also clearly lies on $II_B$, so it is $X$. Similarly, $Y$ is the center of negative homothety of $\omega$ and $\omega_C$. Since $E$ is the "bottom" point (closest to $BC)$ of $\omega_B$, the second intersection of $EX$ with the incircle would be the "top" point of the incircle, since it is the image of $E$ under a negative homothety at $X$ sending $\omega_B$ to $\omega$. Similarly, $FY$ also passes through the "top" point of the incircle, as desired.

Notice $P$ is the insimilicenter of $\omega_B$ and $\omega_C$ by its definition. Then by insimilicenter monge on $\omega_B,\omega_C$ and the incircle of $\triangle ABC,$ we get $X$ is the insimilicenter of $\omega_B$ and the incircle, and similarly $Y$ is the insimilicenter of $\omega_C$ and the incircle. Then by homothety, $EX$ passes through the top of the incircle, and similarly so does $FY,$ which finishes.

Let $Z$ be the top of the incircle $\omega.$ We claim that $EX, FY$ pass through $Z,$ thus finishing the problem. Notice that $P$ is the insimilicenter of $\omega_B, \omega_C,$ and $B$ is the exsimilicenter of $\omega, \omega_B.$ Therefore by Monge d'Alembert $Y$ is the insimilicenter of $\omega_C, \omega.$ Hence, by homothety since $E$ is the bottom of $\omega_B$ it follows that $EX$ passes through $Z.$ By similar reasoning $FY$ passes through $Z,$ so we are done. QED