Generalization: In a triangle $ABC$, $D,E,F$ are arbitrary points chosen on segments $BC,CA,AB$ and $I$ is a random point in the interior of triangle $ABC$. The circumcircles of triangles $AEF,BFD,CDE$ intersect lines $AI,BI,CI$ at points $K,L,M$. Then 4 points $K,L,M,I$ are concyclic. It's easy to prove that the three circumcircles of triangles $AEF,BFD,CDE$ concurrent at a point $N$, and by some simple angle chasing, we can prove that 5 points $K,L,M,N,I$ are concyclic.

it's easy by generalization of the theorem of Napoleon. In the triangle, its angles are $ \frac{A+B}{2} $, $ \frac{B+C}{2} $, $ \frac{C+A}{2} $.

Facts that may help It is not hard to see that $FL=DL,DM=EM,EK=FK.$So applying Ptolemy in quadrilaterals $BDLF,DMEC,FKEA$ gives $FL(BD+BF)=BL*FD$ or $BL=\frac{FL*AC}{FD}$ and similar relations for $CM,AK$.Also see that sine rule in $\triangle FLB$ gives $\frac{FL}{FD}=\frac{1}{2cos\frac{B}{2}}$ and hence $BL=2Rsin\frac{B}{2}$.Analogously $AK=2Rsin\frac{A}{2},CM=2Rsin\frac{C}{2}$.Thus we obtain $BI*BL=AI*AK=CI*CM$.I want some nice approach after this.(without just bashing).

Dear Mathlinkers, in order to have a synthetic proof of the result first presented, can we locate D, E and F? Sincerely Jean-Louis

no ideas? Sincerely Jean-Louis

Dear Mathlinkers, I was thinking, if I am not wrong, to the Nagel triangle of ABC... Can you confirm? Sincerely Jean-Louis

Dear jayme, Points $D,E,F$ is not necessarily fixed(if I understand you correctly).See IMO 2012 Shortlist G6 in page 34. Kind Regards, Arab

Dear Mathlinkers, finally, the five points K, L, M, N and I lie on the Mannheim's circle... See http://perso.orange.fr/jl.ayme vol. 2 Les cercles de Morley..., Mannheim... p. 6-9. Sincerely Jean-Louis

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(23cm); 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 = 0.68, xmax = 23.68, ymin = 0.96, ymax = 12.7; /* image dimensions */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); yaxis(ymin, ymax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((11.48,12.26)--(2.86,3.32)); draw((2.86,3.32)--(18.52,3.98)); draw((11.48,12.26)--(18.52,3.98)); draw(circle((11,9.21), 3.08)); draw(circle((14.36,4.93), 4.26)); draw(circle((6.52,4.9), 3.99)); draw((11.48,12.26)--(15.76,5.78)); draw((2.86,3.32)--(15.76,5.78)); draw((18.52,3.98)--(15.76,5.78)); draw((11.81,8.35)--(15.76,5.78)); /* dots and labels */ dot((11.48,12.26),dotstyle); label("$A$", (11.56,12.38), NE * labelscalefactor); dot((2.86,3.32),dotstyle); label("$B$", (2.94,3.44), NE * labelscalefactor); dot((18.52,3.98),dotstyle); label("$C$", (18.6,4.1), NE * labelscalefactor); dot((10.31,3.63),dotstyle); label("$D$", (10.4,3.76), NE * labelscalefactor); dot((14.08,9.19),dotstyle); label("$E$", (14.16,9.32), NE * labelscalefactor); dot((7.97,8.62),dotstyle); label("$F$", (8.06,8.74), NE * labelscalefactor); dot((15.76,5.78),dotstyle); label("$I$", (15.84,5.9), NE * labelscalefactor); dot((11.81,8.35),dotstyle); label("$M$", (11.9,8.48), NE * labelscalefactor); dot((13.99,8.45),dotstyle); label("$K$", (14.08,8.58), NE * labelscalefactor); dot((10.51,4.78),dotstyle); label("$L$", (10.6,4.9), NE * labelscalefactor); dot((10.29,6.2),dotstyle); label("$N$", (10.38,6.32), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] I will prove the generalization posed by hqdhftw. See obviously the circles $\odot (AEF),\odot (BFD), \odot (CDE)$ concur by Miquel. See there is a configuration problem. So WLOG let $I$ be inside $\odot (CDE)$. Now we see that $\measuredangle{NMC}=\measuredangle{NEC}=\pi -\measuredangle{AEN}=\measuredangle{AFN}=\pi -\measuredangle{BFN}=\measuredangle{BLN}=\pi-\measuredangle{ILN}$ and thus $IMNL$ is concyclic. Also $\pi-\measuredangle{IKN}=\measuredangle{AKN}=\measuredangle{AEN}=\pi-\measuredangle{CMN}\implies \measuredangle{CMN}=\measuredangle{NKI}$ and hence $I,K,M,N$ are concyclic. And this ends the proof since $\odot(IMN)$ is a unique circle. Thus $I,K,L,M,N$ lie on a circle. $\blacksquare$

Hey ! Does that mean all those sum of BD, BF et cetera are useless ? That's bluffing

Yup , i solved the generalization in the exam , but quite a while i was thinking whether there is a flaw in the process or not . But it was right henceforth.