Dear Mathlinklers they are concurrent at the incenter of DEF... Sincerely Jean-Louis

$\rightarrow M=$The midpoint of arc$EF$ which does not contain $D$. $\rightarrow \triangle{AMF} \cong \triangle{AME} \Rightarrow \widehat{MFA}=\widehat{MEA}$. (a) $\rightarrow AF$ is tangent to the incircle $\Rightarrow \widehat{MEF}=\widehat{MFA}$. (b) $\rightarrow ME=MF \Rightarrow \widehat{MEF}=\widehat{MFE}$ (c) $\Rightarrow$ Combining (a),(b),(c) yields that $M \equiv I_1$. Similarly we can prove that $I_2$ and $I_3$ are the midpoints of arc$DF$ and arc$DE$. $\Longrightarrow I_1D,I_2E,I_3F$ are concurrent at the incenter of $\triangle{DEF} \blacksquare$.

Or you can basically observe that $\widehat{FI_1E}=\pi-\widehat{FDE}$, follows that $I_1$ is on the $(DEF)$ circle, but $I_1F=I_1E$, so $I_1$ is the midpoint of the arc $EF$, $DI_1$ being a bisector in $\Delta DEF$...

my proof = LEMMA = In any triangle $ABC$ let $D,E,F$ be points where incentre of triangle $ABC$ touches $BC,CA,AB$ and let $I$ be incentre of triangle $AFE$ than $ A,I,D$ are collinear. PROOF= join $ EI ,FI,AI,ID$ let $ EI , FI $ meet $ AB,AC $ AT $ Z,Y$ let $ ID$ meet $EF$ at $X$ we have $ \angle AIF = 90 + \angle (AEF/2) = 135- \angle (A/4) $ now by steiner lehmus theorem $ EZ = FY$ as $ AF = AE $ also $ \angle ZIF = \angle YEI $ thus , $ FZYE$ is cyclic quad. since $ \angle ZFE = \angle YEF $ thus $ FZYE $ is isosceles trapezium $\Longrightarrow\ ZF =EY$ thus , $ FI = IE$ and hence $ AZ = AY $ and $ \angle IZF = \angle IYE = 90 $ thus $\angle FIX = \angle EIX = \angle FID = 45 +\ angle (A/4) $ thus $ \angle AIF + \angle FID =180 $ hence we proved our lemma. now in our main proof = thus by above lemma in triangle $ABC$ $ I1D , I2E,I3F$ are concurrent by ceva's theorem since we have $ A,I1,D $ and $ B,I2,E$ and $ C,I3,F$ collinear and also we have $ AF=FE , BD=BF , CD=CE $ hence $ (AF/FB) ( BD/DC)(CE/EA) =1 $ and hence $ I1D , I2E, I3F $ are concurrent. hence proved.

Very easy problem. It's easy to see that the points $ I_1,I_2,I_3 $ are lies on incircle.Hence by Steinbart's theorem we must prove that $ AI_1,BI_2,CI_3 $ are concurrent, which is obviously true.

Aditya21, are you sure ?

i dont get you sardor, what you want to say? i didnt know this theorem , hence i tried to prove the lemma.. is my proof of lemma correct?

It is easy to see that $AFE$,$BFD$ and $CDE$ are isosceles . So $ I_1E = I_1F $ Similarly for $I _2$ and $I_3$ Now $\angle EDF= 90^0 - \frac{A}{2} $ .and $ EI_1F = 90^0 + \frac{A}{2} $ $\Longrightarrow I_1 $ lies on $\odot DEF$ .Similarly for $I_2$ and $I_3$ too.Now it follows from these two $I_1D$,$I_2E$ and $I_3F$ are angle bisectors of $\triangle DEF $ .Hnece they concurr at the incentre of $\triangle DEF $.

Aditya, angle IZF is not 90. Let me put it in another way. I is the incentre of triangle AEF, so AI bisects angle EAF = angle BAC. SO AI is the angle bisector of ABC. But I am sure you are aware that D is not the point where the angle bisector meets BC. So your solution is flawed.

Dear Mathlinkers, for the Steinbart-Rabinowitz's theorem, you can see http://jl.ayme.pagesperso-orange.fr/Docs/Les%20points%20de%20Steinbart%20et%20de%20Rabinowitz.pdf Sincerely Jean-Louis

Nice (but easy) problem! $\angle FI_1E=90^{\circ}+\angle \frac{A}2$ and $\angle FDE=90^{\circ}-\angle \frac{A}2 \implies I_1EDF$ is cyclic. Similarly, $I_2DFE$ and $I_3EFD$ are also cyclic, or $I_1,I_2,I_3,D,E,F$ lie on circle with center $I$. Now, $EI_1=FI_2 \implies DI_1$ bisects $\angle EDF$. Similarly, $EI_2$ and $FI_3$ are also the angle bisectors. Hence they concur at incenter of $\Delta DEF$ .

Dear Mathlinkers, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=616655 Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=616655 Sincerely Jean-Louis What does this mean??

Sorry, I haven't the date of this problem..... Jean-Louis

simple angle chasing leads that a i1 and i are collinear by symmetry all concurr at point i which is incenter of traingle abc

Here is a brief solution sketch: Quick angle chase gives us that $I_1,I_2,I_3$ al lie on the incircle of $\triangle ABC$. Now, $\angle I_1FE=\angle I_1DE$, and $\angle I_1EF=\angle I_1DF$. Since $\angle I_1EF=\angle I_1FE$, we have that $I_1D$ is the internal angle bisector of $\angle FDE$. By symmetry we have $I_2E$ is the angle bisector of $\angle DEF$, and that $I_3F$ is the angle bisector of $\angle DFE$. Thus, these lines should meet at the incenter of $\triangle DEF$. We are done!