This is easy,Obviously $AD,BE,CF$ are concurrent by jacobi theorem Let $S=AD\cap BE\cap CF$ then the common radical axis of $\odot (AID),\odot (BIE),\odot (CIF)$ is the isogonal conjugate of the conic through $A,B,C,I,S$

Fumiko wrote: This is easy,Obviously $AD,BE,CF$ are concurrent by jacobi theorem Let $S=AD\cap BE\cap CF$ then the common radical axis of $\odot (AID),\odot (BIE),\odot (CIF)$ is the isogonal conjugate of the conic through $A,B,C,I,S$ You should post the proof to your last claim.

TinaSprout wrote: Fumiko wrote: This is easy,Obviously $AD,BE,CF$ are concurrent by jacobi theorem Let $S=AD\cap BE\cap CF$ then the common radical axis of $\odot (AID),\odot (BIE),\odot (CIF)$ is the isogonal conjugate of the conic through $A,B,C,I,S$ You should post the proof to your last claim. SKETCH: Let $AD\cap BE\cap CF=S$ and let $S'$ be the isogonal conjugate of $S$ wrt $\Delta ABC$ and $I_AI_BI_C$ be the excentral triangle and let $L=\odot (CIF)\cap \odot (BIE)$ ,now it is not hard to show that $L\in \odot (AI_BF),\odot (AI_CE)$ ,now by $\sqrt{bc}$ inversion at $A$ this problem reduces to showing that $I_CE,I_BF,I_AS'$ are concurrent which is not hard ,also it is easy to show that $\Delta I_AI_BI_C,\Delta DEF$ are perspective(actually prespector is $L$) and pedal triangle of $S'$ wrt $\Delta ABC$ is similar to those three mentioned triangle(this is kinda generalization of the fact that pedal triangle of isodynamic point is equilateral)

Let $I_a,I_b,I_c$ be the excenters of $\triangle ABC$ against $A,B,C$ and let $U$ be the 2nd intersection of $\odot(ACE)$ and $\odot(ABF).$ Since $\angle AEC=\angle (UA,UC)$ and $\angle BFA=\angle (UB,UA)$ $\Longrightarrow$ $\angle (UB,UC)=\angle BDC$ $\Longrightarrow$ $U \in \odot(BCD)$ and moreover $A,U,D$ are collinear. Thus if $V$ is the isogonal conjugate of $U$ WRT $\triangle ABC,$ we get $\angle UDB=\angle UCB=\angle ACV$ and $\angle BAD=\angle VAC$ $\Longrightarrow$ $\triangle ABD \sim \triangle AVC$ $\Longrightarrow$ $AB \cdot AC=AD \cdot AV.$ But we know that $AB \cdot AC=AI \cdot AI_a$ $\Longrightarrow$ $AI \cdot AI_a=AD \cdot AV,$ together with $\angle DAI=\angle IAV,$ it follows that $\triangle AIV \sim \triangle ADI_a$ $\Longrightarrow$ $\angle AVI=\angle AI_aD$ and $\angle AIV=\angle ADI_a.$ Hence if $X \equiv IV \cap DI_a,$ it follows that $XAVI_a$ and $XAID$ are concylic $\Longrightarrow$ $X \in \odot(AID)$ and $IV \cdot IX=IA \cdot II_a=IB\cdot II_b=IC \cdot II_c.$ Similarly $\odot(BIE)$ and $\odot(CIF)$ cut $IV$ again at the same point $X$ and the conclusion follows.

That's nice problem dear Telv. Following the solution of Luis, we can see this nice assertion. The circle $(IDA),$ $(IBE)$ and $(ICF)$ are coxial with radical axis is $\ell.$ If we change incenter by excenters, I see that The circles $(I_aAD),$ $(I_aBE)$ and $(I_aCF)$ are coaxial with with radical axis is $\ell_a.$ Define similarly the lines $\ell_b$ and $\ell_c.$ a) Then four lines $\ell_a,$ $\ell_b,$ $\ell_c$ and $\ell$ are concurrent at point $P.$ b) The circles $(PAD),$ $(PBE)$ and $(PCF)$ are also coaxial.

Official solution : Lemma : Given a $ \triangle ABC $ with orthic triangle $ \triangle DEF. $ Let $ X, Y, Z $ be the point such that $$ \measuredangle YDE = \measuredangle FDZ, \measuredangle ZEF = \measuredangle DEX, \measuredangle XFD = \measuredangle EFY. $$Then $ AX, BY, CZ $ are concurrent. Proof : Let $ X^* $ be the point such that $ \triangle AEF \cup X \stackrel{-}{\sim} \triangle ABC \cup X^* $ and define $ Y^*, Z^* $ similarly, then by Jacobi's theorem $ \Longrightarrow $ $ AX^*, BY^*, CZ^* $ are concurrent at $ T, $ so the isogonal conjugate of $ T $ WRT $ \triangle ABC $ lies on $ AX, BY, CZ. $ $ \qquad \square $ ____________________________________________________________ Back to the main problem : Let $ P $ be the second intersection of $ \odot (ECA), \odot (FAB), $ then $$ \measuredangle BPC = \measuredangle BPA + \measuredangle APC = \measuredangle BFA + \measuredangle AEC = \measuredangle BCD + \measuredangle DBC = \measuredangle BDC, $$so $ P \in \odot (DBC). $ Furthermore, from $ \measuredangle APD = \measuredangle APB + \measuredangle BPD = \measuredangle AFB + \measuredangle BCD = 0 $ $ \Longrightarrow $ $ P \in AD. $ Similarly, $ P $ lies on $ BE, CF .$ Let $ \triangle I_aI_bI_c $ be the excentral triangle of $ \triangle ABC $ and let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC. $ From Lemma for $ \triangle II_bI_c $ and $ Q, E, F $ we get $ IQ, I_bE, I_cF $ are concurrent. Analogously, $ IQ, I_cF, I_aD $ are concurrent and $ IQ, I_aD, I_bE $ are concurrent, so $ IQ, I_aD, I_bE, I_cF $ are concurrent at $ S. $ From $ \measuredangle ABQ = \measuredangle PBC = \measuredangle ADC $ we get $ \triangle ABQ \stackrel{+}{\sim} \triangle ADC, $ so $ AI \cdot AI_a = AD \cdot AQ $ and $ \triangle AIQ \stackrel{+}{\sim} \triangle ADI_a $ $ \Longrightarrow $ $ \measuredangle AIS = \measuredangle ADS, $ hence $ S \in \odot (AID). $ Similarly, $ S $ lies on $ \odot (BIE), \odot (CIF). $ $ \qquad \blacksquare $ Remark : This problem is a particular case of the following property which can also be proved synthetically : Given a $ \triangle ABC $ and a point $ P. $ Let $ D,E, F $ be the 2nd intersection of $ AP, BP, CP $ with $ \odot (BPC), $ $ \odot (CPA), $ $ \odot (APB), $ resp. and $ T $ be a point on the circular isogonal pivotal cubic of $ \triangle ABC $ through $ P. $ Then $ \odot (ATD), \odot (BTE), \odot (CTF) $ are coaxial and the radical axis of these three circle passes through the isogonal conjugate of $ P $ WRT $ \triangle ABC. $

Attachments:

By Jacobi, $AD,BE,CF$ concur at $X$, and by spiral similarity, $BCDX,ECAX,BFAX$ are cyclic. Invert at $I$. We get a new problem: in $\triangle ABC$ with orthocenter $H$ let $X$ be a point. Let $P=(AHX)\cap(BCX)$, and define $Q,R$ similarly. Then $AP,BQ,CR$ concur. Let $\triangle DEF$ be the orthic triangle. Let $\mathcal I,\mathcal J$ be the circular points at infinity and construct the cubic $\mathcal C$ through $A,B,C,D,E,F,H,\mathcal I,\mathcal J,X$ which is possible since the first $9$ points are cayley bacharach. If $(AHX)\cap\mathcal C=P'$ then $A+H+X+\mathcal I+\mathcal J+P'=0$, but $A+H+D=B+C+D$ so $B+C+X+\mathcal I+\mathcal J+P'=0$, so $P'\in(BCX)$ and $P'=P$. Similarly $Q,R\in\mathcal C$. Now $A+P=B+Q=C+R=-(H+X+\mathcal I+\mathcal J)$, so $AP,BQ,CR$ concur on $\mathcal C$ as desired.