[asy][asy] size(12cm); defaultpen(fontsize(9pt)); pair D = dir(110), E = dir(210), F = dir(-30); pair I = circumcenter(D, E, F); pair K = dir(190), L = dir(10); pair D1 = extension(K, L, E, F); pair E1 = extension(K, L, D, F); pair F1 = extension(K, L, D, E); pair X = intersectionpoints(circumcircle(E, I, F), D1--E1)[1]; pair Y = intersectionpoints(circumcircle(D, I, F), D1--100*E1)[1]; pair Z = intersectionpoints(circumcircle(E, I, D), D1--E1)[0]; pair M = extension(F, X, D, Z), N = extension(E, X, D, Y); draw(D--E--F--cycle, red+linewidth(1)); draw(circumcircle(D, E, F), heavymagenta); draw(circumcircle(D, E, I), green); draw(circumcircle(D, F, I), green); draw(circumcircle(E, F, I), green); draw(F--D1^^D1--Y^^D1--N^^F--M^^D--Z^^E--N^^D--Y, red); pair XX = extension(Z, F, Y, E); draw(F--Z^^E--Y^^D--XX, red+dashed); string[] names = {"$D$", "$E$", "$F$", "$X$", "$Y$", "$Z$", "$I$", "$D_1$", "$M$", "$N$"}; pair[] points = {D, E, F, X, Y, Z, I, D1, M, N}; pair[] ll = {D, E, F, X, Y, Z, I, D1, M, N}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy]Similar to @above, we restate the problem. BarisKoyuncu wrote: Given a triangle $DEF$ with circumcenter $I$. A line $\ell$ passing through $I$ intersects $(FIE), (FID)$ and $(DIE)$ at $X, Y$ and $Z$, respectively. Prove that $DX, EY$ and $FZ$ are concurrent. Let $\ell$ cuts $EF$ at $D_1$. Let $M, N$ be the second intersection of $\odot(DEF)$ with $FX, EX$ respectively. Claim 01. $D_1, M, N$ are collinear. Proof. Observe that $\measuredangle NXI = \measuredangle EXI = \measuredangle EFI = \measuredangle IEF = \measuredangle IXF$. Therefore, the lines $XN$ and $XF$ are reflection of each other with respect to $\ell$. Since $\ell$ is a diameter of $\odot(DEF)$, therefore $(N, F)$ and $(M, E)$ are reflection of each other with respect to $\ell$. Since $D_1 \in EF$ and $D_1 \in \ell$, this means $D_1 \in MN$ also. $\square$ Claim 02. $D,M,Z$ are collinear, as well as $D, N, Y$. Proof. $\measuredangle(DZ, FX) = \measuredangle(DZ, ZI) + \measuredangle(XI, XF) = \measuredangle(DE, EI) + \measuredangle(EI, EF) = \measuredangle(DE, EF)$. Therefore, $FX \cap DZ \in \odot(DEF) \implies FX \cap DZ = M \implies D, M, Z$ are collinear. Similarly, $D, N, Y$ are also collinear. $\square$ Now observe $\bigtriangleup DZY$ and $\bigtriangleup XFE$. The three points $M = DZ \cap XF, N = DY \cap XE, $ and $D_1 = ZY \cap FE$ are collinear. Therefore by Desargues theorem, the two triangles are perspective, so $DX, ZF, $ and $YE$ are concurrent. $\blacksquare$

Here's my solution in the contest: Let $EF,FD,DE$ intersect line $\ell$ at $K,M,N$, respectively. Then, because $AX \perp KI$ and $KX\cdot KI = KE\cdot KF$ by PoP, $AX$ is the pole of $K$ wrt incircle $\omega$. WLOG, assume that radius of $\omega $ is $1$. $IX=a$, $IY=b$, $IZ=c$. Then $IK= \frac{1}{a}$, $IM=\frac{1}{b}$ and $IN= \frac{1}{c}$ Claim: with directed lengths, $$\frac{KY}{KZ}\cdot \frac{MZ}{MX}\cdot \frac{NX}{NY} = 1$$Proof: $KY=KI+IY= b- \frac{1}{a}$, $KZ=KI+IZ = c-\frac{1}{a}$, then $\frac{KY}{KZ}= \frac{ab-1}{ac-1}$. Similarly, $\frac{MZ}{MX} = \frac{bc-1}{ab-1}$ and $\frac{NX}{NY}= \frac{ac-1}{bc-1}$, so our claim is true. Finally, let the lines perpendicular to $\ell$ from $K,M,N$ intersect lines $BC,CA,AB$ at $K',M',N'$ respectively. Then, polar of $K'$ is $DX$ etc. By our claim, $$\frac{K'B}{K'C} \cdot \frac{M'C}{M'A}\cdot \frac{N'A}{N'B}=1$$which means that $K',M',N'$ are collinear, then by La Hire, lines $DX,EY,FZ$ are concurrent.

Clearly, $A$, $I$, $E$, $F$, and $X$ all lie on the circle with diameter $AI$. Since $IE = IF$, we get that lines $XE$ and $XF$ make the same angle with $\ell$. Similarly, lines $YF$ and $YD$ make the same angle with $\ell$, and so do lines $ZD$ and $ZE$ as well. The result now follows by Jacobi's theorem for the degenerate triangle $XYZ$ and points $D$, $E$, and $F$.

BarisKoyuncu wrote: In a triangle $ABC$, the incircle centered at $I$ is tangent to the sides $BC, AC$ and $AB$ at $D, E$ and $F$, respectively. Let $X, Y$ and $Z$ be the feet of the perpendiculars drawn from $A, B$ and $C$ to a line $\ell$ passing through $I$. Prove that $DX, EY$ and $FZ$ are concurrent. Let $X_1$, $Y_1$, $Z_1$ be the intersections of line $\ell$ with lines $EF$, $FD$, $DE$; $\omega$ be the incenter of $\triangle ABC$ Cuz $EF$ is the polar of $A$ respect to $\omega$ and $AX$ $\perp IX_1$ so $AX$ is the polar of $X_1$ respect to $\omega$ Which means the inversion about $\omega$ sends $X$ to $X_1$ Let $f$ be the inversion about $\omega$. Similar, we have : $f$ : $Y \leftrightarrow Y_1$, $Z \leftrightarrow Z_1$. Lead to $f$ : $DX \leftrightarrow (IX_1D)$, $EY \leftrightarrow (IY_1E)$, $FZ \leftrightarrow (IZ_1F)$. Therefore, to prove $DX, EY$ and $FZ$ are concurrent we just need to prove circles $(IX_1D)$, $(IY_1E)$, $(IF_1Z)$ have a another common point Next, I will rephrase this problem again in general case and prove it : $\, \,$"Given a triangle $ABC$ with circumcenter $O$ and $d$ is a random line. Suppose that $d$ insect lines $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively and $P$ is the foot of altitude from $O$ to $d$. Then circles $(PAD)$, $(PBE)$, $(PCF)$ have a common point different to $P$." $\textbf{Proof :}$ Let $U$ be the second intersection of line $AB$ and circle $(PBE)$, $V$ be the second intersection of line $AC$ and circle $(PCF)$ $\hspace{0.5cm}$$\mathcal{P}_\omega (X)$ be the power of a point $X$ respect to a circle $\omega$ First, i will prove that $\frac{\mathcal{P}_{(PBE)} (A)}{\mathcal{P}_{(PCF)} (A)} = \frac{\mathcal{P}_{(PBE)} (D)}{\mathcal{P}_{(PCF)} (D)} $ $(*)$ : $\, \,$Use Menelaus theorem for $\triangle AEF$ with points $D$, $B$, $C$ we have : $\frac{\overline{DE}}{\overline{DF}} .\frac{\overline{CA}}{ \overline{CE}} .\frac{\overline{BF}}{\overline{BA}} = 1 \Leftrightarrow \frac{\overline{DE}}{\overline{DF}} = \frac{\overline{CE}}{ \overline{CA}} .\frac{\overline{BA}}{\overline{BF}}$ $\, \,$Therefore, $(*) \Leftrightarrow \frac{\overline{AB} .\overline{AU}}{\overline{AC} .\overline{AV}} = \frac{\overline{DE} .\overline{DP}}{\overline{DF} .\overline{DP}} \Leftrightarrow \frac{\overline{AU}}{\overline{AV}} = \frac{\overline{CE}}{\overline{BF}} \Leftrightarrow \overline{AU} .\overline{BF} = \overline{AV} .\overline{CE}\Leftrightarrow \overline{AF}.\overline{BF} + \overline{FU} .\overline{BF} = \overline{AE}.\overline{CE} + \overline{EV}.\overline{CE}$ $\, \,$Combine with $\overline{EV}.\overline{EC} = \overline{EP}.\overline{EF} = \mathcal{P}_{(PCF)} (E)$, $\overline{FU}.\overline{FB} = \overline{FP}.\overline{FE} = \mathcal{P}_{(PBE)} (F)$, $\hspace{2.2cm} \overline{AF}.\overline{BF} = \mathcal{P}_{(ABC)} (F) = OA^2 - OF^2$, $\overline{AE}.\overline{CE} = \mathcal{P}_{(ABC)} (E) = OA^2 - OE^2$ $\, \,$We have $(*) \Leftrightarrow OA^2 - OF^2 - \overline{EP}.\overline{EF} = OA^2 - OE^2 - \overline{FP}.\overline{FE} \Leftrightarrow PF^2 - PE^2 = \overline{FE}.(\overline{FP} - \overline{PE}) $, we ez so see this is true In brief, we have $\frac{\mathcal{P}_{(PBE)} (A)}{\mathcal{P}_{(PCF)} (A)} = \frac{\mathcal{P}_{(PBE)} (D)}{\mathcal{P}_{(PCF)} (D)} $ Next, let $\Gamma$ be the circle pass through $A$ and satisfying that three radical axis of circles $\Gamma$, $(PBE)$, $(PCF)$ is a same line $\hspace{1.1cm}$ $\Gamma'$ be the circle pass through $D$ and satisfying that three radical axis of circles $\Gamma'$, $(PBE)$, $(PCF)$ is a same line $\hspace{1.1cm}$$O_1$, $O_1'$, $O_2$, $O_3$ be the circumcenter of circle $\Gamma$, $\Gamma'$, $(PBE)$, $(PCF)$ respectively I will prove that $O_1 \equiv O_1'$ : $\, \,$$\textbf{Lemma 1:}$ $\, \,$Given two circles $\gamma_1$, $\gamma_2$ and a point $A$ in the plane. With $I$, $J$ are circumcenter of $\gamma_1$, $\gamma_2$ respectively, $K$ is the foot of altitude from $A$ to radical axis of $\gamma_1$ and $\gamma_2$ , we have $\mathcal{P}_{\gamma_1} (A) -\mathcal{P}_{\gamma_2} (A) = 2.\overline{JI}.\overline{AK}$ $\, \,$Proof of lemma : $\hspace{0.4cm}$Let $H$ be the intersection of radical axis of $\gamma_1$ and $\gamma_2$ with line $IJ$; $M$ be the midpoint of segment $IJ$ $\hspace{0.9cm}$$i$, $j$ be radius of circles $\gamma_1$, $\gamma_2$ respectively $\hspace{0.4cm}$We have : $\mathcal{P}_{\gamma_1} (A) -\mathcal{P}_{\gamma_2} (A) = AI^2 - i^2 - AJ^2 + j^2 = AI^2 - AJ^2 + HJ^2 - HI^2 $ (Cuz $\mathcal{P}_{\gamma_1} (H) = \mathcal{P}_{\gamma_2} (H)$) $\hspace{4.7cm}$$= (\overrightarrow{AI} + \overrightarrow{AJ})(\overrightarrow{AI} - \overrightarrow{AJ}) + (\overrightarrow{HJ} + \overrightarrow{HI})(\overrightarrow{HJ} - \overrightarrow{HI}) = 2.\overrightarrow{AM}.\overrightarrow{JI} + 2\overrightarrow{HM}.\overrightarrow{IJ} $ $\hspace{4.7cm}$$= 2\overrightarrow{IJ}(\overrightarrow{HM} - \overrightarrow{AM}) = 2\overrightarrow{IJ}.\overrightarrow{HA} = 2.\overline{JI}.\overline{AK}$ From Lemma 1, let $K$ be foot of altituder from $A$ to common radical axis of three circles $\Gamma$, $(PBE)$, $(PCF)$, we see that : $\, \,$$\mathcal{P}_(PBE) (A) = 2.\overline{O_1O_3}.\overline{AK}$; $\mathcal{P}_(PCF) (A) = 2.\overline{O_1O_2}.\overline{AK}$ Lead to $\frac{\mathcal{P}_(PBE) (A)}{\mathcal{P}_(PCF) (A)} = \frac{\overline{O_1O_3}}{\overline{O_1O_2}}$. Similar, $\frac{\mathcal{P}_(PBE) (D)}{\mathcal{P}_(PCF) (D)} = \frac{\overline{O_1'O_3}}{\overline{O_1'O_2}}$ Therefore, $\frac{\overline{O_1O_3}}{\overline{O_1O_2}} = \frac{\overline{O_1'O_3}}{\overline{O_1'O_2}}$. Which means $O_1 \equiv O_1'$, $\Gamma \equiv \Gamma'$ Thus, circles $(PAD)$, $(PBE)$, $(PCF)$ have a common point different to $P$, done

We have $X\in (AIEF), \ Y\in (BIDF), \ Z\in (CIDE)$. Invert from $I$ with radius $ID$. $X^*\in EF,Y^*\in DF,Z^*\in DE$ and $X^*,Y^*,Z^*$ are collinear. New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$. $K$ is an arbitrary point on $BC$ and $KO$ intersects $AB,AC$ at $M,L$ respectively. Then, $(AOK),(BOL),(COM)$ are concurrent on a point other than $O$. Proof: We will use the method of moving points. Let $\phi$ represent the inversion centered at $O$ with radius $OA$. \[f:K\mapsto KO\cap AC=L\mapsto \phi(L)\mapsto \phi(L)B\cap A\phi(K)\mapsto \phi(\phi(L)B\cap A\phi(K))\]\[g: K\mapsto KO\cap AB=M\mapsto \phi(M)\mapsto \phi(M)C\cap A\phi(K)\mapsto \phi(\phi(M)C\cap A\phi(K))\]$max\{deg \ f,deg \ g\}\leq 4$ hence we will study on $4$ cases. $\bullet \ K=B$ $f:B\mapsto BO\cap AC\mapsto B\mapsto B$ and $g:B\mapsto B$ $\bullet \ K=C$ $f:C\mapsto CO\cap AB\mapsto C\mapsto C$ and $g:C \mapsto C$ $\bullet \ K=N$ where $N$ is the midpoint of $BC$. Let $ON$ intersect $AB,AC$ at $X,Y$. Note that $\phi(X)=Y\iff \phi(Y)=X$. Let $P\in AB,Q\in AC$ and $\overline{POQ}\parallel BC$. $f:N\mapsto X\mapsto Y\mapsto A\mapsto A$ and $g:N\mapsto Y\mapsto X\mapsto A\mapsto A$ $\bullet \ K=BC_{\infty}$ $f:BC_{\infty}\mapsto Q\mapsto \phi(\phi(Q)B\cap AO)$ and $g:BC_{\infty}\mapsto P\mapsto \phi(\phi(P)C\cap AO)$ Thus we want to prove that $P^*C,Q^*B,AO$ are concurrent. When we invert back, we want to show that $(BPO),(CQO),AO$ are concurrent. Let $(BOP)\cap AC=S$ and $(CQO)\cap AC=T$.$\angle BSP=90-\angle A$ yields $\angle CBS=90-\angle B$ thus $A'\in BS$ where $A'$ is the antipode of $A$ on $(ABC)$. Similarily we get that $C,A',T$ are collinear. We can see that $(A,S,T,A')$ is an orthogonal system so $B,C,S,T$ are cyclic. We have $\angle QTS=\angle BCA=\angle QPA$ hence $P,Q,T,S$ are cyclic. $Pow(A,(BOPS))=AP.AS=AQ.AT=Pow(A,(COQT))$ thus, $A$ lies on the radical axis of those two circles which gives the desired result.$\blacksquare$

Let $YF\cap ZE=D_1, \ DZ\cap XF=E_1, \ DY\cap EX=F_1, \ YE\cap ZF=P, DX\cap (DEF)=X'$. Claim: $D_1,E_1,F_1\in (DEF)$. Proof: \[\measuredangle ZD_1Y=180-\measuredangle D_1YZ-\measuredangle YZD_1=180-\measuredangle FDI-\measuredangle IDE=180-\measuredangle FDE\]Hence $D_1\in (DEF)$. Similarily $E_1$ and $F_1$ lie on $(DEF)$.$\square$ Claim: $DD_1\perp l$. Proof: Let $l$ intersect $DE,DF$ at $Z^*,Y^*$ respectively. Since $Y^*,Z^*$ are the images of $Y,Z$ under the inversion centered at $I$ with radius $ID$, $D_1$ lies on both $(IZ^*E)$ and $(IY^*F)$. \[\measuredangle IZ^*D_1=\measuredangle IED_1=\measuredangle ED_1I=\measuredangle DZ^*I\]Thus, $IZ^*\equiv l$ is the perpendicular bisector of $DD_1$.$\square$ Apply DDIT on quadrilateral $YEZF$ with involution center $D$. $(\overline{DF_1},\overline{DE_1}),(\overline{DE},\overline{DF}),(\overline{DP},\overline{DD_1})$ is an involution. Let $D_1X\cap (DEF)=Q^*$. \[\measuredangle XQ^*I=\measuredangle ID_1X=\measuredangle XDI\]Which implies $I,X,D,Q^*$ are concyclic. $\measuredangle E_1IF_1=2(180-\measuredangle F_1DE_1)=180-\measuredangle A$ so \[\measuredangle E_1XF_1=\measuredangle FXE=180-\measuredangle A=\measuredangle E_1IF_1\]This yields $I\in (XE_1F_1)$. Inverting from $X$ with radius $\sqrt{-XD.XX'}$ gives $D_1X',EF,E_1F_1$ are concurrent. Thus, $(\overline{DD_1},\overline{DX}),(\overline{DE},\overline{DF}),(\overline{DE_1},\overline{DF_1})$ is an involution. Hence $DX\equiv DP$ which implies $D,X,P$ are collinear as desired.$\blacksquare$