[asy][asy] defaultpen(fontsize(9pt)); size(6cm); pair A,B,C,I,N,I1,D,E,F,M,P,K,T,X,V,V1,A1; A=dir(120); B=dir(210); C=dir(330); I=incenter(A,B,C); N=circumcenter(B,I,C); I1=-I+2N; D=foot(I1,B,C); E=foot(I1,A,B); F=foot(I1,A,C); M=(B+C)/2; P=extension(C,E,B,F); K=-I+2M; T=(A+P)/2; X=(E+F)/2; V=foot(I,B,C); V1=-V+2I; A1=-A+2M; draw(A--B--C--cycle);draw(B--E--F--C); draw(incircle(A,B,C)); draw(incircle(A1,B,C),linewidth(0.2)); draw(arc(circumcenter(D,E,F),circumradius(D,E,F),0,180)); draw(A--P); draw(C--E); draw(B--F); draw(A--X--T); draw(D--T); draw(I--K); draw(P--X); draw(V--V1); draw(B--A1,linewidth(0.3)); draw(C--A1,linewidth(0.3)); dot("$A$" , A , dir(A)); dot("$B$" , B , dir(B)); dot("$C$" , C , dir(30)); dot("$I$" , I , dir(I)); dot("$D$" , D , dir(70)); dot("$E$" , E , dir(180)); dot("$F$" , F , dir(60)); dot("$P$" , P , dir(P)); dot("$M$" , M , dir(60)); dot("$K$" , K , dir(K)); dot("$T$" , T , dir(80)); dot("$X$" , X , dir(X)); dot("$V$" , V , dir(V)); dot("$V'$" , V1 , dir(70)); dot("$A'$" , A1 , dir(A1)); [/asy][/asy] $\measuredangle BNE=\measuredangle EFA=\measuredangle AEN\implies \overline{BN}\parallel\overline{AC}$. Similarly we can get that $\overline{CM}\parallel\overline{AB}$. Let $\overline{BN}\cap\overline{CM}=X$, so we have that $ABXC$ is a parallelogram. Now notice that $$XB+BD=AC+BD=b+(s-b)=s=c+(s-c)=AB+CD=XC+CD$$where $s$ is the semiperimeter of $\Delta ABC$. Hence, $\overline{XD}$ is the $X-$ Nagel Cevian of $\Delta XBC$. So, $D,N,M$ are touchpoints of the $X$- Excircle of $\Delta XBC$ with $\overline{BC},\overline{XB},\overline{XC}$ respectively. Let $I'$ be the incenter of $\Delta XBC$, then notice that $IBCI'$ forms a parallelogram, so if $K$ is the midpoint of $BC$ then $I'$ is the reflection of $I$ over $K$. With all these information we can reformulate the problem as follows. $\textbf{REFORMULATION:-}$ $ABC$ be a triangle with $\Delta DEF$ as the $A-$ Extouch triangle, $I$ as the incenter and $M$ as the midpoint of $BC$. Let $K$ be the reflection of $I$ over $M$ and $X$ be the midpoint of $EF$ also let $\overline{CE}$ meet $\overline{BF}$ at $P$. Then $P,K,X$ are collinear. Let the incircle of $\Delta ABC$ touch $\overline{BC}$ at $V$ and $V'$ be the reflection of $V$ over $I$. It's well known that $A,V',D$ are collinear and $MV=MD$. Hence, $\overline{IM}\parallel\overline{AD}$. Let $T$ be the midpoint of $AP$, hence, $\overline{TM}$ is the Gauss-Bondenmiller Line of the complete quadrilateral $\mathbf{Q}\equiv\{\overline{AB},\overline{BF},\overline{CE},\overline{AC}\}\implies T,M,X$ are collinear also clearly $A,I,X$ are collinear. If $\overline{XP}\cap\overline{IM}=K'$. Then $\frac{\overline{AT}}{\overline{TP}}=\frac{\overline{MI}}{\overline{MK'}}=1\implies IM=IK'=IK\implies K'\equiv K$. Hence, $P,K,X$ are collinear. $\quad\blacksquare$

$\newline \underline{\textbf{Solution:}}$ Let $G$ be the midpoint of $\overline{MN}$. We first state a lemma Lemma: The incircle of $\triangle ABC$ is tangent to $ BC, AC,$ and $AB$ at $D, E, F$, respectively. Le $M,N$ be the midpoints of $BC,AC$ respectively. Ray $BI$ meets lines $EF$ at $K$. Then, $BK \perp CK$ $\newline \underline{\textbf{Proof:}}$ We shall look at the $\triangle FBK$ .As we know $\angle FBK = \frac{1}{2}\angle ABC $, or $\angle FBK = \frac{1}{2}\angle B$. Also we know that $\angle KFB = 90 + \frac{1}{2}\angle A$,we get this $\frac{1}{2}\angle A$ from the cyclic $\square AFIE$ Thus we can get $\angle FKB = 180-(\angle FBK + \angle KFB) = 90 - \frac{1}{2} \angle B - \frac{1}{2} \angle A = 90 - 90 + \frac{1}{2} \angle C = \angle C $ But as we know $\angle ECI = \frac{1}{2} \angle C = \angle FKB = \angle EKI $,and thus the $\square CKEI$ is cyclic. $\newline \rule{\textwidth}{0.5pt}$ Now, let $Z$ be the orthocenter of $\Delta IBC$ and, let $S$ and $T$ be the feet of altitude from $C$ and $B$ to $\overline{CI}$ and $\overline{BI}$ respectively. Then, $Z$ is the circumcenter of $\Delta DMN$ $\newline \underline{\textbf{Proof:}}$ Let $N'$ be a point $\overline{ST}$ such that $\angle ZN'B = 90^{\circ}.$ Then, there exists a circle through quadrilateral $ZN'BDT.$ Therefore, $\angle BDN' = \angle BTN' = \angle BTD ($ as orthocenter is incenter of its orthic triangle $) = \angle BN'D.$ Which means that $BN' = BD$ or $N' \equiv N$ and hence $BS$ is perpendicular bisector of $DN.$ Similarly, $CT$ is perpendicular bisector of $DM$ and both the bisectors concur at $Z$. Hence $Z$ is the circumcenter of $\triangle DMN$ $\newline \rule{\textwidth}{0.5pt} \newline$ Let $U$ be a point on $\overline{ST}$ such that $\overline{DU} \perp \overline{ST}$. If $V$ is the midpoint of $\overline{DU},$ we prove that $\overline{P-V-I-G},$ where $\overline{ZG} \perp \overline{ST}$ from above. $\newline \rule{\textwidth}{0.5pt}$ Lemma; In triangle $ABC$ with incenter $I$ and $A-$excenter $I_A,$ let $D$ and $T$ be the incircle and $A-$excircle tangency point with $BC.$ If $M$ is the midpoint of the $A-$altitude, then $M, I, T$ collinear and $M, D, I_{A}$ collinear. $\newline \underline{\textbf{Proof: }}$ We define $AM\cap BC = K, AE\cap KD = X$ and $AD \cap \odot(I_{A}) = Y$ such that $XY$ is diameter of $I_{A}$ Then, $\newline$ i) Notice that $AK\parallel ED$ and $X=AE\cap KD$. Consider the homothety at $X$ sending $ED$ to $AK$. This sends $I$ to $M$, so $X,I,M$ are collinear. $\newline$ ii) Notice that $AK\parallel XY$ and $D=KX\cap AY$. Consider the homothety at $D$ sending $XY$ to $KA$. This sends $I_A$ to $M$, so $D,I_A,M$ are collinear. $\blacksquare$ $\newline$ We know that $Z$ is the $D-$excenter of $\Delta DTS$ and therefore, $G$ is the $D-$extouch point. $I$ is the orthocenter of $\Delta ZBC,$ Hence by above lemma, $\overline{ V-I-G.}$ $\newline \rule{\textwidth}{0.5pt} \newline$ Let $W$ and $A_{1}$ be the midpoints of $\overline{DN}$ and $\overline{DM}$ respectively. Now, there exists a homothety that sends $WA_{1} \leftrightarrow NM$ or $W \leftrightarrow N$ and $A_{1} \leftrightarrow M.$ Therefore, $BM \cap CN = P \leftrightarrow BW \cap CA_{1} = P'$ (say). Hence, it suffices to prove that $\overline{ P'-V-I}.$ By applying Pappus' theorem to points ${B, W, S}$ and ${C, A_{1}, T}$ in that order to get $\overline{(BA_{1} \cap CW)-(WT \cap A_{1}S)-(BT \cap CS)}$ but $BA_{1} \cap CW = P', WT \cap A_{1}S = X$(say), $BT \cap CS = I.$ Therefore we have that $\overline{ P'-X- I}$ $\newline \rule{\textwidth}{0.5pt}$ We eventually come to a point where it suffices to prove that $\overline{V-I-X}.$ We know that $VI \cap ST = G$ and $AG \parallel DU$ (perpendicular to same line). Let $ZD \cap ST = B_{1}$ and it is well-known that $(Z, I; B_{1}, D) = -1.$ Hence, $$(Z, I; B_{1}, D) \stackrel{G}{=} (\infty_{DU}, \overline{GI} \cap \overline{DU}; U, D) = (U, D; \overline{GI}\cap \overline{DU}, \infty_{DU}) = -1$$Which essentially means that $ \overline{GI} \cap \overline{DU}$ is the midpoint of $\overline{DU}.$ Hence, $\overline{GI} \cap \overline{DU} \equiv V$ or $\overline{V-I-X}$ which completes our problem

[asy][asy] import olympiad; size(7cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(120), B = dir(225), C = dir(315), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), M = 2*foot(C,E,F)-E, N = 2*foot(B,E,F)-F, Q = (M+N)/2, P = extension(B,M,C,N), K = foot(I,B,M), L = foot(I,C,N); draw(A--B--C--A, linewidth(0.8)); draw(incircle(A,B,C), gray); draw(N--B--M--N--C--M--N); draw(P--Q, dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$E$", E, dir(75)); dot("$F$", F, dir(135)); dot("$I$", I, dir(135)); dot("$M$", M, dir(75)); dot("$N$", N, dir(135)); dot("$P$", P, dir(105)); dot("$Q$", Q, dir(120)); [/asy][/asy] We proceed with barycentric coordinates wrt $ABC$. We know $I = (a:b:c)$, $E = (s-c:0:s-a)$, and $F = (s-b:s-a:0)$. Note that $\overline{BN} \parallel \overline{AC}$ by construction, which means $N = (k,1,-k)$ for some $k$. Also $N$ lies on $\overline{EF}$, so we have \begin{align*} 0 &= \begin{vmatrix} k & 1 & -k \\ s-c & 0 & s-a \\ s-b & s-a & 0 \end{vmatrix} \\ &= (s-a)[-k(s-a)-k(s-c)+(s-b)] \\ \implies k(2s-a-c) &= s-b \\ \implies k &= \frac{s-b}{b},\end{align*}which means $$N = \left(\frac{s-b}{b},1,\frac{b-s}{b}\right) = (s-b:b:b-s).$$Similarly $$M = \left(\frac{s-c}{c}, \frac{c-s}{c}, 1\right) = (s-c:c-s:c).$$Intersecting cevians $\overline{BM}$ and $\overline{CN}$ gives $$P = ((s-b)(s-c):b(s-c):c(s-b)).$$Let $Q$ be the midpoint of $\overline{MN}$ so that we want to show $P$, $I$, $Q$ collinear. We have $$P = \frac{M+N}{2} = (s(b+c)-2bc:b(2c-s):c(2b-s)).$$It remains to verify \begin{align*}0 &= \begin{vmatrix} s(b+c)-2bc & b(2c-s) & c(2b-s) \\ (s-b)(s-c) & b(s-c) & c(s-b) \\ a & b & c\end{vmatrix} \\ \iff 0 &= \begin{vmatrix} s(b+c)-2bc & 2c-s & 2b-s \\ (s-b)(s-c) & s-c & s-b \\ a & 1 & 1\end{vmatrix} \\ &= [s(b+c)-2bc](b-c) + (s-b)(s-c)(2b-2c) + a[(2c-s)(s-b) - (2b-s)(s-c)] \\ &= (b-c)[s(b+c)-2bc+2(s-b)(s-c)] + a(cs-bs) \\ &= s(b-c)(2s-b-c-a),\end{align*}which is true, as desired.

[asy][asy] import olympiad; size(15cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.5)); dotfactor *= 1.5; pair A=(-1.22,4.39); pair B=(-3.64,-0.95); pair C=(4.7,-0.91); pair E=(0.817,2.566); pair F=(-2.349,1.9); pair G=(-2.956,0.559); pair H=(1.914,1.584); pair I=(-0.521,1.071); pair M=(6.851,3.837); pair N=(-5.971,1.137); pair P=(-1.11,0.204); pair X=(0.44,2.487); dot("$A$", A,dir(120)); dot("$B$", B,SW); dot("$C$", C,SE); dot("$E$",E,E); dot("$F$",F,NW); dot("$G$", G,dir(135)); dot("$H$", H,E); dot("$I$",I,dir(315)); dot("$M$", M,E); dot("$N$", N,E); dot("$P$", P,S); dot("$X$",X,dir(120)); draw(A--B--C--cycle); draw(B--M); draw(C--N); draw(M--N); draw(G--H); draw(I--E); draw(I--F); draw(A--I); draw(B--N); draw(C--M); draw(P--X, dashed); [/asy][/asy] Let $\angle BAC=A,\angle ABC=B$ and $\angle BCA=C$. Let the $A-\text{mixtilinear incircle}$ touches $AB,AC$ at $G,H$ respectively. It can be seen that $AF = AE$ and $AG = AH$ which means $EF\parallel GH$. We know $I$ lies on $GH$. So, $I$ is midpoint of $GH$ and $AI\perp GH$. Let $X$ be midpoint of $MN$. Claim: $G,H$ lies on $CN, BM$ respectively. Proof: We first show that $H\in BM$ and $G\in CN$ follows similarly. Since, $AF=AE$, $CE=CM$ and $\angle AEF=\angle CEM$ we get $\angle FAE=\angle ECM$ which implies $AB\parallel CM$. Let $BH$ meet $EF$ at $M'$. So, it suffices to show that $AB\parallel CM'$. Since, $EF\parallel GH$ we have $\triangle BFM'\sim \triangle BGH$ which gives $\frac{BG}{GF}=\frac{BH}{HM'}$. It now suffices to show that $$\frac{BG}{GF}=\frac{AH}{CH}$$as then $\triangle CHM'\sim \triangle AHB$ which implies $AB\parallel CM'$. It is well known that $\angle AIB=90^\circ +(C/2)$. So by law of sines in $\triangle AIB$ we get $$\frac{AI}{\sin (B/2)}=\frac{AB}{\sin(90^\circ +(C/2))}\Longrightarrow AB=AI\frac{\cos(C/2)}{\sin(B/2)}.$$Similarly, $AC=AI\frac{\cos(B/2)}{\sin(C/2)}$. Now, \begin{align*} \frac{BG}{GF} &=\frac{AB-AG}{AG-AF}\\& =\frac{AI\left(\frac{\cos(C/2)}{\sin(B/2)}-\frac{1}{\cos(A/2)}\right)}{AI\left(\frac{1}{\cos(A/2)}-\cos(A/2)\right)}\\ &= \frac{\frac{\cos(C/2)\cos(A/2)-\sin(B/2)}{\sin(B/2)\cos(A/2)}}{\frac{\sin^2(A/2)}{\cos(A/2)}} \\&= \frac{\cos(C/2)\cos(A/2)-\sin(B/2)}{\sin(B/2)\sin^2(A/2)} \end{align*} also, \begin{align*} \frac{AH}{CH} &=\frac{AH}{AC-AH} \\&=\frac{\frac{AI}{\cos (A/2)}}{AI\left( \frac{\cos(B/2)}{\sin(C/2)}-\frac{1}{\cos(A/2)} \right)}\\&= \frac{\frac{1}{\cos(A/2)}}{\frac{\cos(B/2)\cos(A/2)-\sin(C/2)}{\sin(C/2)\cos(A/2)}}\\&= \frac{\sin(C/2)}{\cos(B/2)\cos(A/2)-\sin(C/2)} \end{align*} So, it now suffices to show that \begin{align*} &\frac{\cos(C/2)\cos(A/2)-\sin(B/2)}{\sin(B/2)\sin^2(A/2)} = \frac{\sin(C/2)}{\cos(B/2)\cos(A/2)-\sin(C/2)} \\ \iff & (\cos(B/2)\cos(A/2)-\sin(C/2))(\cos(C/2)\cos(A/2)-\sin(B/2))=\sin^2(A/2)\sin(B/2)\sin(C/2)\\ \iff & \cos(B/2)\cos^2(A/2)\cos(C/2)-\frac{1}{2}\sin(B)\cos(A/2)-\frac{1}{2}\sin(C)\cos(A/2)+\sin(B/2)\sin(C/2) = \sin^2(A/2)\sin(B/2)\sin(C/2) \\ \iff & \cos(B/2)\cos(C/2)\cos^2(A/2)+\sin(B/2)\sin(C/2)\cos^2(A/2)=\frac{1}{2}\cos(A/2)(\sin(B)+\sin(C)) \\ \iff & \frac{1}{2}\cos^2(A/2)\left(\cos\left(\frac{B-C}{2}\right)+\cos\left(\frac{B+C}{2}\right)+\cos\left(\frac{B-C}{2}\right)-\cos\left(\frac{B+C}{2}\right)\right)=\frac{1}{2}\cos(A/2)(\sin(B)+\sin(C)) \\ \iff & \sin(B)+\sin(C)=2\cos(A/2)\cos\left(\frac{B-C}{2}\right)=2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right) \end{align*}The last line is trivially true and hence the claim is proved. $\square$ As $I$ is midpoint of $GH$, by homothety with center $P$ sending $GH$ to $MN$ (this holds good since $GH\parallel MN$), $I$ is sent to $X$. Hence, $P,I,X$ are collinear and we are done. $\square$

Dunno why yall bruddas be overcomplicating Let $X, Y$ be $\ell \cap AB, AC$, respectively, where $\ell$ is the line parallel to $EF$ through $I$. It suffices to show $X \in CN$ and then similarly $Y \in BM$, and we would be done by homothety. We length chase - let $a, b, c$ be the $A, B, C$ tangent lengths, respectively. Note that by right triangle lengths\[r^2 = IF^2 = AF \cdot FX \implies FX = \frac{r^2}{a} = \frac{bc}{a + b + c} \implies \frac{AX}{BX} = \frac{a + FX}{b - FX} = \frac{a + c}{b}.\]Next, let $CN \cap AB = X'$. Note that $BNF$ is isosceles with vertex $B$ hence $\angle NBF = \angle EAF$ thus $BN \parallel AE$ and so it follows that $\triangle X'BN \sim \triangle X'CA$ so by similar triangle lengths\[\frac{AX'}{BX'} = \frac{AC}{BN} = \frac{a + c}{b}\]hence $X = X'$, as desired. Similarly $Y = Y'$ and we are done. $\blacksquare$

Let $a=BC$, $b=CA$, $c=AB$, and let $s$ denote the semiperimeter. Let $X$, $Y$ be the tangency points between the $A$-mixtillinear incircle and $\overline{AC}$, $\overline{AB}$. Then $X$, $I$, $Y$ lie on a line parallel to $\overline{EF}$; moreover, since the inverse of $X$ under $\sqrt{bc}$-inversion is the tangency point between the $A$-excircle and $\overline{AB}$, we know $AX=bc/s$. We are given that $\overline{CM}\parallel\overline{AB}$ and $\overline{BN}\parallel\overline{AC}$. Observe that \[\frac{AX}{XC}=\frac{bc/s}{b-bc/s}=\frac c{s-c}=\frac{AB}{CE}=\frac{AB}{CM},\]so $B$, $X$, $M$ are collinear. Similarly $C$, $Y$, $N$ collinear, so since $I$ is the midpoint of $\overline{XY}$, line $PI$ bisects $\overline{MN}$ by homothety.

Let the line through \(I\) parallel to \(\overline{EF}\) intersect \(\overline{AC}\) and \(\overline{AB}\) at \(X\) and \(Y\). Let lines \(BI\) and \(EF\) intersect at \(K\), so by Iran lemma, \(K\) lies on the \(C\)-midline. Since \(\overline{CM}\parallel\overline{AB}\), it follows that \(K\) is the midpoint of \(\overline{FM}\). Since \(I\) is the midpoint of \(\overline{XY}\) and \(\overline{XY}\parallel\overline{EF}\), lines \(YF\), \(IK\), \(XM\) concur by homothety, so \(B\), \(X\), \(M\) are collinear. Similarly \(C\), \(Y\), \(N\) are collinear, and by homothety at \(P\), line \(PI\) bisects \(\overline{MN}\). [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=heavyred; pen pri2=lightred; pen sec=lightblue; pen tri=purple+pink; pen qua=fuchsia+pink+dashed; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,I,EE,F,M,NN,P,X,Y,Q,MA,K; A=dir(110); B=dir(225); C=dir(315); I=incenter(A,B,C); EE=foot(I,C,A); F=foot(I,A,B); M=extension(EE,F,C,C+A-B); NN=extension(EE,F,B,B+A-C); P=extension(B,M,C,NN); X=extension(A,C,B,M); Y=extension(A,B,C,NN); Q=(M+NN)/2; MA=(B+C)/2; K=extension(B,I,EE,F); draw(B--K--MA,qua); draw(X--Y,pri2); draw(B--M,tri);draw(C--NN,tri); fill(B--F--NN--cycle,sfil); fill(C--EE--M--cycle,sfil); draw(C--M--NN--B,sec); filldraw(incircle(A,B,C),fil,pri2); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(M_A\)",MA,S); dot("\(E\)",EE,dir(60)); dot("\(F\)",F,NW); dot("\(M\)",M,NE); dot("\(N\)",NN,NW); dot("\(I\)",I,dir(120)); dot("\(P\)",P,S); dot("\(K\)",K,NW); dot("\(X\)",X,dir(-15)); dot("\(Y\)",Y,dir(220)); [/asy][/asy]

My solution for this nice problem.

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Thanks for your interest in this problem. An inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CM=CE$ and $BN=BF$. Let $Q_a$ be the midpoint of $MN$. Define similarly the points $Q_b$ and $Q_c$. i) Prove that triangle $Q_aQ_bQ_c$ and $ABC$ have the same centroid. ii) Lines $AQ_a$, $BQ_b$, $CQ_c$ are concurrent at $Q$. iii) Prove that line $IQ$ goes through NPC center of triangle $Q_aQ_bQ_c$.

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Another inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CN=CE$ and $BM=BF$. Lines $BN$ and $CM$ meet at point $P$. $A$-mixtilinear excircle touches $CA$, $AB$ at $K$, $L$, respectively. $BK$ meets $CL$ at $Q$. Prove that rays $AP$ and $AQ$ are isogonal in $\angle BAC$.

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buratinogigle wrote: Another inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CN=CE$ and $BM=BF$. Lines $BN$ and $CM$ meet at point $P$. $A$-mixtilinear excircle touches $CA$, $AB$ at $K$, $L$, respectively. $BK$ meets $CL$ at $Q$. Prove that rays $AP$ and $AQ$ are isogonal in $\angle BAC$. Taking a $\sqrt{AB\cdot AC}$ inversion with a flip around the angle bisector of $\angle ABC$, it just suffices to show that $\overline{AP}$ is the radical axis of $\odot(AEB),\odot(AFC)$. Let $\overline{AP}\cap\overline{BC}=X$. Define $\overline{CM}\cap\overline{AB}=U$ and $\overline{BN}\cap\overline{AC}=V$. By Ceva, $\frac{XC}{XB}=\frac{AU}{AB}\cdot\frac{CV}{VA}=\frac{b(s-c)}{c(s-b)} (\bigstar)$. Let $\odot(ABE)=\omega_1$ and $\odot(ACF)=\omega_2$. Define $f(\bullet)=\mathcal{P}_{\omega_1}(\bullet)-\mathcal{P}_{\omega_2}(\bullet)$. Using Corollay 5.5 from here and $(\bigstar)$ we get $f(X)=0\implies\mathcal{P}_{\omega_1}X=\mathcal{P}_{\omega_2}X\implies AP$ is the radical axis of $\odot(ABE)$ and $\odot(ACF)$, now invert back to get the desired conclusion. $\blacksquare$

buratinogigle wrote: Another inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CN=CE$ and $BM=BF$. Lines $BN$ and $CM$ meet at point $P$. $A$-mixtilinear excircle touches $CA$, $AB$ at $K$, $L$, respectively. $BK$ meets $CL$ at $Q$. Prove that rays $AP$ and $AQ$ are isogonal in $\angle BAC$. My proof. $P$ has coordinates from #4 $$P = ((s-b)(s-c),b(s-c),c(s-b)).$$Isogonal conjugate of $P$ $$P^*=\left(\frac{a^2}{(s-b)(s-c)},\frac{b}{s-c},\frac{c}{s-b}\right)\sim(a^2,b(s-b),c(s-c)).$$$AQ$ goes through Mittenpunkt point $$Mt=(a(s-a),b(s-b),c(s-c)).$$Thus, $A$, $P^*$, $Q$ are collinear.

Another inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CN=CE$ and $BM=BF$. Lines $BN$ and $CM$ meet at point $P$. $K$ is the orthocenter of triangle $DEF$. $L$ is the reflection of $K$ in $I$. Prove that rays $DP$ and $DL$ are isogonal in $\angle EDF$.

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buratinogigle wrote: Thanks for your interest in this problem. An inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CM=CE$ and $BN=BF$. Let $Q_a$ be the midpoint of $MN$. Define similarly the points $Q_b$ and $Q_c$. i) Prove that triangle $Q_aQ_bQ_c$ and $ABC$ have the same centroid. ii) Lines $AQ_a$, $BQ_b$, $CQ_c$ are concurrent at $Q$. iii) Prove that line $IQ$ goes through NPC center of triangle $Q_aQ_bQ_c$. Using the calculation of barycentric coordinate in #4. Rewrite the coordinates of $Q$ in homogeneous, we have $$Q_a= \left(\frac{b^{2} + c^{2} + ab + ac - 2bc}{4bc}, \frac{-a - b + 3c}{4c}, \frac{-a + 3b - c}{4b} \right).$$$$Q_b= \left( \frac{-a - b + 3c}{4c}, \frac{a^{2} + c^{2} + ab - 2ac + bc}{4ac}, \frac{3a - b - c}{4a} \right).$$$$Q_c= \left( \frac{-a + 3b - c}{4b}, \frac{3a - b - c}{4a}, \frac{a^{2} + b^{2} - 2ab + ac + bc}{4ab} \right).$$Easily seen, $AQ_a$, $BQ_b$, $CQ_c$ are concurrent at $$Q=\left(\frac{a}{b+c-3a},\frac{b}{c+a-3b},\frac{c}{a+b-3c}\right).$$Using formula of distance, we have $$q_a^2=Q_bQ_c^2=\left(a - b - c \right)\frac{a^{4} - 3a^{3}b - 3a^{3}c - 7a^{2}b^{2} + 26a^{2}bc - 7a^{2}c^{2} - ab^{3} - 3ab^{2}c - 3abc^{2} - ac^{3} + 2b^{4} - 4b^{2}c^{2} + 2c^{4}}{16cba}.$$$$q_b^2=Q_cQ_a^2=\left(b - a - c \right)\frac{b^{4} - 3b^{3}a - 3b^{3}c - 7b^{2}a^{2} + 26b^{2}ac - 7b^{2}c^{2} - ba^{3} - 3ba^{2}c - 3bac^{2} - bc^{3} + 2a^{4} - 4a^{2}c^{2} + 2c^{4}}{16cab}.$$$$q_c^2=Q_aQ_b^2=\left(c - b - a \right)\frac{c^{4} - 3c^{3}b - 3c^{3}a - 7c^{2}b^{2} + 26c^{2}ba - 7c^{2}a^{2} - cb^{3} - 3cb^{2}a - 3cba^{2} - ca^{3} + 2b^{4} - 4b^{2}a^{2} + 2a^{4}}{16abc}.$$Using barycentric coordinates of NPC center, we have $$N_q=\left(q_a^2(q_b^2+q_c^2)-(q_b^2-q_c^2)^2,\ ,\ \right)$$$$\sim(a^{5}b + a^{5}c + 3a^{4}b^{2} - 24a^{4}bc + 3a^{4}c^{2} + 2a^{3}b^{3} + 18a^{3}b^{2}c + 18a^{3}bc^{2} + 2a^{3}c^{3} - 2a^{2}b^{4} + 48a^{2}b^{3}c - 92a^{2}b^{2}c^{2} + 48a^{2}bc^{3} - 2a^{2}c^{4} - 3ab^{5} + 5ab^{4}c - 34ab^{3}c^{2} - 34ab^{2}c^{3} + 5abc^{4} - 3ac^{5} - b^{6} + 9b^{4}c^{2} + 16b^{3}c^{3} + 9b^{2}c^{4} - c^{6},\ ,\ ).$$Since $I=\left(a,b,c\right)$, we can check $$\det\begin{pmatrix}Q\\N_q\\I\end{pmatrix}=0.$$Done!

solutions

amar_04 wrote: The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CM=CE$ and $BN=BF$. Lines $BM$ and $CN$ meet at point $P$. Prove that $PI$ bisects segment $MN$. An addition. Let $D$ be the projection of $I$ on $BC$. Prove that reflection of $A$ in $I$ lies on the line $PD$.

buratinogigle wrote: Another inspiration. The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CN=CE$ and $BM=BF$. Lines $BN$ and $CM$ meet at point $P$. $K$ is the orthocenter of triangle $DEF$. $L$ is the reflection of $K$ in $I$. Prove that rays $DP$ and $DL$ are isogonal in $\angle EDF$. buratinogigle wrote: amar_04 wrote: The incircle of triangle $ABC$ centered at $I$ touches $CA,AB$ at points $E,F$ respectively. Let points $M,N$ of line $EF$ be such that $CM=CE$ and $BN=BF$. Lines $BM$ and $CN$ meet at point $P$. Prove that $PI$ bisects segment $MN$. An addition. Let $D$ be the projection of $I$ on $BC$. Prove that reflection of $A$ in $I$ lies on the line $PD$. So we can reformulate above problem Let $ABC$ be a triangle with circumcircle $(O)$ and de Longchamps point $L$ (reflection of orthocenter in circumcenter). Let $T$ be the intersection of tangents at $B$ and $C$ of $(O)$. Let $S$ be the reflection of $T$ in $O$. Prove that lines $AS$ and $AL$ are isogonal in angle $\angle BAC$.