It suffices to prove that $\angle{IDG}=\angle{AEI}$. Taking the excenter we have to prove that the triangles $AFI_a$ and $AIE$ are similar. But this is easy because it is enough to show that $\frac{AF}{AI_a}=\frac{AI}{AE}$. But from the similarity of ABF,AEC we have that $\frac{AE}{AC}=\frac{AB}{AF}$. So we have to prove that $AI\cdot AI_a=AB\cdot AC$ which is clearly true.

I have a different aproach.. Suppose EI cuts $ \Gamma $ at K. Let the parallel from I to BC cut AF to P. Then $ AKPI $ is cyclic (because AK is antiparallel to IP or by simple angle chase). Now we prove that D,P,K are collinear:From the cyclic $ AKPI $ $ \widehat{AKP}= \widehat{PID}= \widehat{PIB} + \widehat{BID}= \widehat{B} + \widehat{A}/2 $. But $ \widehat{AKD}= \widehat{ABD}= \widehat{ABC} + \widehat{CBD}= \widehat{B} + \widehat{A}/2 $. If line DPK cuts BC at Q, it suffices to prove that $ IQ \parallel AF $ since then, $ PIQF $ will be parallelogram and G the intersection of DK and IE. Since $ \widehat{IAP}= \widehat{IKP} $ and we want to show $ \widehat{IAP}= \widehat{DIQ} $ it is enough $ DI^2= DQ \cdot DK $ But DI=DB (well-known fact) so we have to prove $ DB^2=DQ \cdot DK$ which is obvious from similar triangles $ DBQ,DKB $ since arcs BD and DC are equal. QED Image not found Note: the condition $ \angle BAF=\angle CAE <\frac{1}2\angle BAC $ it is not necessary.

Let $L$ be the second intersection of $IE$ and $\Gamma$. I will prove that $G, D, L$ are collinear Let $M$ be a point on the prolongation of $AD$ ($D$ is between $A$ and $M$) such that $\angle{IBM}=90$ $\angle{IBC}=\frac{1}{2}\angle{B}$ and $\angle{CBD}=\frac{1}{2}\angle{A}$. Then $\angle{DBM}=\frac{1}{2}\angle{C}$, $\angle{BIM}=\frac{1}{2}\angle{A}+\frac{1}{2}\angle{B}$. Then $\angle{BMI}=\frac{1}{2}\angle{C}$. Then $BD=DM$. Then $BD$ bisects $IM$ because $\angle{IBM}=90$. Then $ID=DM$ (I believe all this results are well-known) $\triangle{ABM}$ is similar to $\triangle{AIC}$. Then $\frac{AM}{AC}=\frac{AB}{AI}$ $\triangle{AEC}$ is similar to $\triangle{ABF}$. Then $\frac{AE}{AC}=\frac{AB}{AF}$ Then we have $\frac{AM}{AE}=\frac{AF}{AI}$. But $\angle{FAM}=\angle{IAE}$. Then $\triangle{AFM}$ and $\triangle{AIE}$ are similar. Then $\angle{FMA}=\angle{IEA}=\angle{LDA}$ Then $LD$ and $FM$ are parallel. But $ID=DM$ then $LD$ bisects $FI$. Then $G, D, L$ are collinear.

I just want to add that there is another way of continuing my solution after we got that $ID=DM$ which I believe is well-known. This is the continuation: Let $AI=i, AD=l, AB=c, AC=b$. Then $AM=2l-i$ $\triangle{ABM}$ is similar to $\triangle{AIC}$. Then $\frac{AM}{AB}=\frac{AC}{AI}$. Then $\frac{2l-i}{c}=\frac{b}{i}$ Then $(2l-i)i = bc$. Then $il-i^2=bc-il$ (*) Let $N$ the intersection of $LD$ and $AF$. $\frac{AN}{AD}=\frac{AI}{AE}$ because $\triangle{AND}$ and $\triangle{AIE}$ are similar. Then $AN=\frac{il}{AE}$ $\frac{AF}{AB}=\frac{AC}{AE}$ because $\triangle{ABF}$ and $\triangle{AEC}$ are similar. Then $AF=\frac{bc}{AE}$ Then $FN=AF-AN=\frac{bc-il}{AE}$ Let $G'$ the intersection of $LD$ and $AF$. Then $\frac{IG'}{G'F}\frac{FN}{AN}\frac{AD}{ID}=1$ by Menelaus Then $\frac{IG'}{G'F}=\frac{ID}{AD}\frac{AN}{FN}=\frac{il(l-i)}{(bc-il)l}=\frac{il-i^2}{bc-il}=1$ by (*) Then $G'=G$, and then $D, G, L$ are collinear. This problem can also be solved automatically using complex numbers. The solution is long and painful to write, but I will try to post it soon

Another solution: We need to prove that <GMI = <IEA. It is well known (and it can be proved easily) that the midpoint D of the arc BC is the center of the circumcircle U of the triangle BIC. Let T be the symmetric of I with respect to D, which is the intersection of the line AD and the circle U, then FT//MG and <FTI = <GMI, so we need <FTI = <IEA, or its enough to prove that triangles AFT and AIE are similliar, or equivalently AI/AE = AF/AT <=> AI*AT = AE*AF (1). From the similliar triangles ABF and AEC we have AE/AC = AB/AF <=> AE*AF = AB*AC. So we need AI*AT = AB*AC <=> AI/AB = AC/AT, which is true because the triangles ABI, ATC are similliar (<BAI = <TAC = A/2, <ABI = <ABC/2 = <ADC/2 = <ATC)!

Problem with a lot of points on a circumference calls for Pascal theorem. Historical note: he was 13 y.o. when he discovered it, probably solving something similar to this Problem #2 at French National math olympiad years ago. Mr. T

We show that $\triangle{AFI_a}$ and $\triangle{AIE}$ are similar. Then we have $\angle{AEI}=\angle{AI_a F}=\angle{ADG}$ and we're done. To show that, inverse the plane with regard to $A$ with the radius $\sqrt{bc}$ where $b=AC,c=AB$. Then we have another figure which can be also obtained by reflecting the original figure. Note that $E,F$ are mapped to $F,E$ resp. Hence $AE \cdot AF = bc = AI \cdot AI_a$, which implies directly that $\triangle{AFI_a}$ and $\triangle{AIE}$ are similar, as desired.

orl wrote: Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. Let $D'$ be a point on $AI$ such that $DI = DD'$. Notice that $\angle IDG = \angle AD'F$. So we only have to prove that $\angle AD'F = \angle AEI$. We have $\triangle ABF\sim\triangle AEC$, therefore $AB\cdot AC = AE\cdot AF\quad (1)$. $\triangle ABI\sim\triangle AD'C$ $\Longrightarrow AB\cdot AC = AI\cdot AD'\quad (2)$. Combine $(1)$ and $(2)$, we have $AE\cdot AF = AI\cdot AD'$, i.e. $\dfrac{AF}{AD'}=\dfrac{AI}{AE}$. On the other hand, $\angle FAD' = \angle IAE$, so $\triangle FAD'\sim\triangle IAE$. It follows $\angle AD'F=\angle AEI$, which completes our solution.

To add a little algebraic point of view... The points where $EI$ and $DG$ meet $\Gamma$ are functions of $\angle AEI$ and $\angle ADG$, which we must prove are equal. Let $DG$ meet $AF$ at $T$. Since $\angle TAD = \angle IAE$, we must prove that $TAD$ and $IAE$ are congruent. To do this, we need a spiral similarity between $ATI$ and $ADE$. On the other hand, naming $\theta = \angle BAF$, we see that $\angle AFC = \angle ADE = \angle ABC + \theta$. Thus, if $\{P\} = FC \cap AD$, we have $AFP$ similar to $ADE$, and it remains to prove $AFP$ is similar to $ATI$, or $TI \parallel FP$. If this happens, let ${Q} = BC \cap DG$. Since $G = \frac{F + I}{2}$, $TIQF$ must be a parallelogram. Reversely, if $QI \parallel AF$, we are going to have that parallelogram. In other words, we need to prove $QIP$ is similar to $FAP$. We prove this happens in the degenerate cases $F = B$ and $F = C$. When $F$ varies on (in? at?) $BC$, $G$ varies linearly, and so does $P$, because it is a mean of $G$ and $D$, with weights that depend only on the ratio $IP/PD$. So our linearity ends the argument. If $F = B$ (draw another figure!), however, $BID$ is isosceles, so that $DG$ is a simmetry axis, which leads to $\angle PIQ = \angle DIQ = \angle DBQ = \frac{1}{2}\angle BAC = \angle PAB$. And here is the similarity!

Let $M$ be the midpoint of $AI$. We want to show that $\angle GDM = \angle IEA$, that is, $\triangle MGD \sim \triangle AIE$. $MG$ is parallel to $AF$, so $\angle GMD = \angle FAD = \angle IAE$. Hence, $\triangle MGD \sim \triangle AIE$ if and only if $\frac{MG}{MD} = \frac{AI}{AE}$. $MG = \frac{AF}{2}$, so this is equivalent to $2 AI \cdot MD = AE \cdot AF$. We claim that $AE \cdot AF$ is fixed. It is then sufficient to show that the result is true for some choice of $E$ and $F$ (namely, when $E=C$ and $F=B$), as it would imply that $2 AI \cdot MD = AE \cdot AF$ for some choices of $E$ and $F$, and thus for all choices of $E$ and $F$. Showing that $AE \cdot AF$ is not difficult. $\angle ABF = \angle AEC$ and $\angle BAF = \angle CAE$, so $\triangle ABF \sim \triangle AEC$, so $AE \cdot AF = AB \cdot AC$. That the result is true when $F=B$ and $E=C$ is not difficult to show either. In this case, $G$ is the midpoint of the base of isosceles $\triangle IBD$, so $DG$ is the bisector of $\angle BDA$, so it meets $\Gamma$ on the midpoint of minor arc $AB$. On the other hand, $CI$ trivially meets $\Gamma$ on the midpoint of minor arc $AB$ as well, so our proof is complete.

Let $EI$ intersect $\Gamma$ again at $P$. We'll prove that $DP$ meets $FI$ at its midpoint. Let $DP$ cut $AF$ and $BC$ at $X$ and $Y$, resp. Notice that $APXI$ is cyclic, because $\angle XAI = \angle DAE = \angle XPI$. We'll see that $XFYI$ is a parallelogram. Lemma 1: $XI$ is parallel to $FY$. Proof: As $APXI$ is cyclic, we have $\angle PAI = \angle DXI$. But $\angle PAI$ is equal to the angle between $DP$ and the tangent to $\Gamma$ through $D$. Then, it follows that $XI$ and that tangent are parallel. But as $D$ is the midpoint of arc $BC$, $XI$ must be parallel to $BC$ too, and the lemma follows. Lemma 2: $D$ is the circumcenter of $BIC$. Proof: As $D$ is the midpoint of arc $BC$, we have $DB = DC$. Now, let the circle with center $D$ which passes through $B$ and $C$ meet $AD$ at $I'$. We know that $\angle I'CB = \angle BDI'/2 = \angle BCA/2$. Thus $I'$ is also on the bisector of $\angle C$, so $I'$ is the incenter, and the lemma follows. Lemma 3: $FX$ is parallel to $YI$. Proof: Consider the inversion with center $D$ and radius $DB$. It maps $BC$ into $\Gamma$, and, by lemma 2, fixes $I$. Now, as $Y$ is mapped to $P$ and $I$ is fixed, we have $\angle YID = \angle DPI$. But we know that $XPAI$ is cyclic, thus $\angle YID = \angle XAI$, and the result follows. Now, by lemmas 1 and 3, we get that $XFYI$ is a parallelogram. Thus $DP$ meets $FI$ at its midpoint, which is what we wanted to prove.

orl wrote: Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. 2010 IMO Problem 2 was proposed by Tai Wai Ming (2008 Hong Kong IMO team member) and Wang Chongli

wenxin wrote: orl wrote: Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. 2010 IMO Problem 2 was proposed by Tai Wai Ming (2008 Hong Kong IMO team member) and Wang Chongli Oh really !? Tai Wai Ming was my teammate but we have not kept in contact since IMO 2008

silouan wrote: It suffices to prove that $\angle{IDG}=\angle{AEI}$. Taking the excenter we have to prove that the triangles $AFI_a$ and $AIE$ are similar. But this is easy because it is enough to show that $\frac{AF}{AI_a}=\frac{AI}{AE}$. But from the similarity of ABF,AEC we have that $\frac{AE}{AC}=\frac{AB}{AF}$. So we have to prove that $AI\cdot AI_a=AB\cdot AC$ which is clearly true. The most short and nice solution of this problem! I have the same, but it took of me about 2,5 hours to get it.

I do not think anyone has used Menelaos yet (probably because it does not yield a very short solution) Let $DG$ intersect $AF$ and $BC$ at $M$ and $N$, respectively. Applying Menelaos' Theorem to $AFI$, we obtain $\frac{AM}{MF}\frac{FG}{GI}\frac{ID}{DA}=1\Longleftrightarrow\frac{AM}{MF}=\frac{AD}{DI}$ since $G$ is the midpoint of $[IF]$. Now, straightforward computation on the line containing $A$, $I$, $D$ and $J$, the foot of the angular bisector through $A$, yields $AD\cdot IJ=DI\cdot AI$ - it is basically the power-of-a-point condition $AJ\cdot DJ=IJ\cdot(2DI-IJ)$ rearranged, so $\frac{AM}{MF}=\frac{AI}{IJ}$ Thus $MI\parallel BC$. Now let $H$ and $H'$ the second points of intersection of $EI$ and $DG$, respectively, with $\Gamma$. Observe that $\widehat{BJA}=\widehat{DCA}=\widehat{DEA}$, so $AFJ$ and $ADE$ are similar, whence $\widehat{ADE}=\widehat{AFJ}=\widehat{AMI}$. Now $\widehat{AHI}=\widehat{ADE}$, so $\widehat{AHI}=\widehat{AMI}$, so $H$ lies on the circumcircle of $AMI$. Also, $\widehat{AH'I}=180^\circ-\widehat{AED}=\widehat{FJA}=\widehat{MIA}$, so $H'$ lies on the circumcircle of $AMI$ as well. Thus $H$ and $H'$ both lie on the circumcircle of $AMI$ and on $\Gamma$. But neither of these points can coincide with $A$ (since $D\not=E$), so $H=H'$, and we are done.

Here's an outline: Lemma: $AE*AF-AI*AD=2Rr$. Proof: We can compute all four side-lengths in terms of parameters of the triangle and $\angle{BAF}=\angle{CAE}$. Specifically, $AE$ and $AD$ are chords, so we can get both them in terms of $R$, and we can get $AI$ in terms of $r$. Then, compute all of the trig expressions (really, it's not too bad), using the fact that $2[ABC]=ac\sin{B}=r(a+b+c)$, the law of sines, and in the end the sum-to-product formula. From here, we want to let CI meet the circumcircle at P, then let PD meet AF and FI at X and G', respectively. Note that AIE ~ AXD. Menelaus on AIF reduces the problem to AX/XF=AD/DI, which is equivalent to AI/AE=XF/DI, and AI*DI=AE*XF. But AI*DI is the (negative) power of I with respect to the circumcircle, which is 2Rr, and we can find that AE*XF=AE*AF-AI*AD by the fact that AX*AE=AI*AD. Then, we're done by the lemma.

Is valid that SP.SP=SC.SC=SA.SB , so the triangles ABP , ΑSP are similar. Hence χ=α=mod arc ΒΚ/2 =arc BK/2(more simple) (1) Since angl.SPC=angl.SCP or (χ+ψ)-χ=ω ,we take arc( ML/2+ BC/2-BK/2)= arc ( BC/2+MB/2) Hence arc ML/2=arcMK/2 or MK=ML

why triangles $AFI_a$ and $AIE$ are similar??

Solved with CyclicISLscelesTrapezoid, v4913, crazyeyemoody907, CircleInvert, rjiangbz, DottedCaculator, awesomehuman, razmath, ApraTrip. Let $I_a$ be the $A-$ excenter, so that $\triangle AFI_a\sim \triangle AIE$. Then if $X = EI\cap DG$, we have $$\angle DXE = \angle (I_aF, EI) = \angle (I_aA, EA) = \angle DAE$$ as desired.

We prove G is the midpoint given the lines concur. Let $K = HD \cap AF$, $J = KD \cap BC$, and $P = AD \cap BC$ Claim: $KI \parallel BC$ Proof: Since $\angle KHI = \angle KAI$, HKAI is cyclic. Now see that by using the inscribed angle theorem, $\angle AHD = \frac12 \widehat{ACD} = 180 - (\angle DAC + \angle ACB) = 180 - \angle JPA$ so $HAJP$ is cyclic. This implies $\angle AIK = \angle APF$ proving $KI \parallel BC$. By Menelaus on $\triangle AIF$, the midpoint condition is equivalent to $\frac{FK}{KA} \cdot \frac{AD}{DI} = 1$ or $\frac{AK}{AF} = \frac{AD}{DI}$. This is true because $\frac{AK}{KF} = \frac{AI}{IP} = \frac{AC}{CP}$ and $\frac{AD}{DI} = \frac{AD}{DB}$ by Fact 5 but those are equal as $\triangle ABD $ is similar to $\triangle APC$

This problem can be solved using the method of projective moving points. The first projectivity goes from line $BC$ to the pencil of lines through $I$, to the line parallel to $BC$ halfway to $I$, to the pencil of lines through $D$ to the circle $ABC$. This is the line $DG$. The next projectivity sends $BC$ to the circumcircle of $ABC$ via $\sqrt{bc}$ inversion, then it is projected onto itself through $I$. This is the line $IE$. To prove that they always intersect, we need to show that these are the same projectivity, so we just need to check $3$ cases. First: set $F$ as the point at infinity of line $BC$. The first projectivity sends $F$ to the infinity point of the halfway line, and then it is projected onto $(ABC)$ through $D$, which is $D$ itself. The second projectivity sends $F$ to $A$ through $\sqrt{bc}$ inversion and then to $D$ through $I$. Second: set $F$ as the point on $BC$ that lies in the internal angle bisector of $A$. The first projectivity trivially sends $F$ to $A$ and the second one sends $F$ to $D$ through $\sqrt{bc}$ inversion and then to $A$ through $I$. Third: set $F$ as the contact point of the $A-$excircle. The first projectivity sends $F$ to a point on the perpendicular bisector of $BC$ then $D$ projects it onto the midpoint of arc $BAC$. Then the second projectivity sends $F$ to the contact point of the mixtillinear circle through the inversion, and I sends it to the same midpoint of the arc.

Very enjoyable and nice configuration in general. Let $I_A$ be the $A$-excenter, and $H = \overline{EI} \cap (ABC)$. It suffices to show that $\overline{HD} \parallel \overline{FI_A}$ by homothety reasons (then it passes through $G$.) On the other hand, notice that $AIE$ and $AFI_A$ are spirally similar as $AE \cdot AF = AB \cdot AC = AI \cdot AI_A$. This implies $L = \overline{FI_A} \cap \overline{IE}$ lies on $(AIF)$. So $$\measuredangle IHD = \measuredangle EAD = \measuredangle DAF = \measuredangle ILF,$$implying the result.

Wasted a couple hours trying to find a synthetic without constructing anything, i seriously need to get better at this; also i am NOT going to type in latex properly Construct the A-excenter I_a, and let $H=DG\cdot IE$; we want to prove $AI_aF=ADG=AEI$, and since we know $FAI_a=EAI_a$, we're motivated to try to prove that $AFI_a\similar AIE$. Indeed, we have $$BAI=I_aAC,ABI=IBC=II_aC\implies ABI\similar AI_aC\implies AB\cdot AC=AI\cdot AI_a\implies ABF\similar AEC\implies AE\cdot AF=AB\cdot AC=AI\cdot AI_a,$$upon which dividing yields SAS similarity and we're done. $\blacksquare$

Assume $AB\le{}AC$. The case where $AB>AC$ is similar. Let $K'$ be the intersection of $\overline{EI}$ and $\Gamma$, let $X$ be the intersection of $\overline{AF}$ and $\overline{K'D}$, and let $Y$ be the intersection of $\overline{K'D}$ and $\overline{BC}$. We have that $\angle{}DK'E=\angle{}DAE=\angle{}FAD$, so $K'AIX$ is cyclic. This means that $\angle{}AXI=\angle{}AK'E=\angle{}ABE=\angle{}AFC$ since $\triangle{}ABE\sim\triangle{}AFC$, so $\overline{XI}\parallel\overline{BC}$. It is therefore sufficient to prove that $\overline{AF}\parallel\overline{IY}$. Let the circumcircle of $\triangle{}AK'I$ intersect $\overline{AB}$ and $\overline{AC}$ at points $M$ and $N$ and let the intersection of $\overline{MN}$ and $\overline{AI}$ be $W$. We have that $\angle{}K'MN=180^\circ-\angle{}K'AC=\angle{}K'BC$ and $\angle{}K'NM=\angle{}K'AB=\angle{}K'CB$, so $\triangle{}K'MN\sim\triangle{}K'BC$. The spiral similarity mapping $\triangle{}K'MN$ to $\triangle{}K'BC$ maps $I$ to $D$ and maps $A$ to a point $Z$. We claim that $\overline{DZ}\parallel\overline{AF}$. We will do this by showing that $\angle{}AWN=\angle{}AFC$. Notice that \begin{align*} \angle{}AWN&=\angle{}AMN+\angle{}BAD\\ &=\angle{}AMI-\angle{}IMN+\angle{}BAD\\ &=\angle{}AMI-\angle{}IAN+\angle{}BAD\\ &=\angle{}AMI-\frac{\angle{}CAB}{2}+\frac{\angle{}CAB}{2}\\ &=\angle{}AMI\\ &=\angle{}AXI\\ &=\angle{}AFC \end{align*}since $K'AIX$ is cyclic and $\overline{XI}\parallel\overline{BC}$, proving the claim. It is then sufficient that $\angle{}YID=\angle{}IDZ$. However, since $\triangle{}K'AI\sim\triangle{}K'ZD$, we have that \begin{align*} \angle{}IDZ&=\angle{}K'DZ-\angle{}K'DA\\ &=\angle{}K'IA-\angle{}K'DA\\ &=\angle{}EK'D, \end{align*}so it is sufficient that $\triangle{}DYI\sim\triangle{}DIK'$, or $DY\cdot{}DK'=DI^2$. However, by the Incenter-Excenter Lemma, we have that $DI^2=DB^2$, so it is sufficient to prove that $DY\cdot{}DK'=DB^2$, or $\triangle{}DYB\sim\triangle{}DBK'$. Therefore, it is sufficient that $\angle{}YBD=\angle{}BK'D$, but this is true since $\angle{}BK'D=\angle{}BAD=\angle{}CAD=\angle{}CBD=\angle{}YBD$.

first off, let $AF$ intersect $\Gamma$ at $Q$, let $EI$ intersect $\Gamma$ at $P$, let $PD$ intersect $AQ$ at $L$, let $BC$ intersect $PD$ at $M$, and let $PD$ intersect $IF$ at $J$ it is obvious that $QD=ED$ and $BQ=EC$, so $QE$ is parallel to $BC$ our problem is equivalent to proving $DP$ bisects $IF$, or $J$ is the midpoint of $IF$ we have $QPD=EPD=QAD=DAE$, $PQA=PDA=PEA$, $DAB=DCB=DAC$, $PQA=PDA=PEA$, so $PQL$, $PDI$, $ADL$ and $AEI$ are all similar from this, we can get that $APLI$ are concyclic, and $LI$ is parallel to $QE$ and $BC$ we can then get that $FJM$ and $LJI$ (to be finished later)

Let $X$ and $P$ be the second intersections between $\Gamma$ and $EI$, $AF$ respectively. Let $Y=DX \cap AP$, $M=DX \cap IF$. We need to show that $M$ is the midpoint of $IF$. Note that $BC$, $EP$, and the tangent to $\Gamma$ and $D$ are parallel. By Pascal's theorem on $ADDXEP$, $IY$ is parallel to $BC$. We proceed by length chasing. Let $Q=IY \cap DF$, $H=AD \cap BC$. By Ceva's theorem on the cevians $DM$, $FA$, $IQ$ of $\Delta DFI$, intersecting at $Y$, $IM=MF \iff \frac{QF}{QD}=\frac{AI}{AD} \iff \frac{DH}{HI}=\frac{DI}{AI}$. For convenience let $\angle BAC = 2\alpha$, $\angle ABC = 2\beta$, $\angle ACB = 2\gamma$. $$\frac{DH}{HI}=\frac{\frac{BH}{\sin 2\gamma} \cdot \sin \alpha}{\frac{BH}{\sin (90^{\circ}-\gamma)} \cdot \sin \beta}=\frac{\sin \alpha}{2\sin\beta \sin\gamma}.$$$$\frac{DI}{AI}=\frac{BD}{\frac{AB}{\sin(90^{\circ}-\gamma)}\cdot \sin \beta}=\frac{BD}{AB} \cdot \frac{\cos \gamma}{\sin \beta}=\frac{\sin \alpha}{\sin 2\gamma} \cdot \frac{\cos \gamma}{\sin \beta}=\frac{\sin \alpha}{2\sin\beta \sin\gamma}.$$This concludes the proof. $\square$

Delete the $<\tfrac{1}{2} \angle BAC$ condition. We prove a stronger version of the problem statement with MMP. Define $K_1=\Gamma \cap \overline{EI}$ and $K_2=\Gamma \cap \overline{DG}.$ Fix $\triangle ABC$ and animate $F$ linearly along $\overline{BC}.$ Then $I,D$ are fixed and $E,K_1,K_2,G$ vary as a function of $F.$ Note $F \mapsto E$ is a projective map since rotations preserve cross ratios. Furthermore, $E \mapsto K_1$ is a projective map, so $F \mapsto K_1$ is projective. On the other hand, by homothety about $I$ with ratio $1/2,$ we know $G$ moves linearly. Taking perspectivity at $D,$ the map $G \mapsto K_2$ is projective. It suffices to show these maps coincide for $3$ values of $F.$ This is not hard to do. Check $F=B, F=C,$ and $F=\overline{AI} \cap \overline{BC}.$

Let $X = \overline{EI} \cap \overline{DG}$ and denote the $A$-excenter of $\triangle ABC$ as $I_A$ It suffices to show $\angle ADX = \angle AEX$. It is very clear that $\overline{DG} \parallel \overline{I_AF}$. Then, observe that \begin{align*} \angle ADX = \angle AEX & \iff \angle AI_AF = \angle AEI \\ & \iff \triangle AEI \sim \triangle AI_AF \\ & \iff \frac{AE}{AI} = \frac{AI_A}{AF} \\ & \iff AE \cdot AF = AI \cdot AI_A. \end{align*} However, $\sqrt{bc}$ inversion swaps $I, I_A$ and $E, F$ so the conclusion follows immediately. $\square$ Remarks: The inversion is not necessary to prove $AE \cdot AF = AB \cdot AC$; in fact, I realized that similar triangles are way easier to prove in retrospect. A more rigorous way to show that $E$ and $F$ are swapped is to realize that they lie on isogonal conjugates in $\angle BAC$ and that $\overline{BC}$ and $(ABC)$ are swapped under the inversion.

orl wrote: Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. Proposed by Tai Wai Ming and Wang Chongli, Hong Kong Let $I_A$ be the $A$ excenter then: By symetric invertion with center $A$ we get that $AF\cdot AE=AI\cdot AI_A\Rightarrow \frac{AF}{AI}=\frac{AI_A}{AE}$ and with $\angle FAI_A=\angle IAE$ we get that $AFI_A\approx AIE$ Now since $D$ is the midpoint of $II_AS$ and $G$ the midpoint of $IF$ we get that: $\angle IDG=\angle II_AF=\angle AEI\Rightarrow EI\cap DG\varepsilon (ABC)$

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Let $X$ be the intersection of $EI$ with $(ABC)$, we will show that $X-G-D$ are collinear. 1) Show that $XAIS$ cyclic. 2) Show that $IQSF$ is a parallelogram, this implies that $XD $ bisects segment $IF$ which means that $G$ Lies on $XD$. proof: By shooting lemma and some ratios.

Let $EI\cap \Gamma=T, AF\cap \Gamma=K$. If we show that $DT$ bisects $IF$, we were done. Let $DT\cap AL=L$. Since $\angle{LTI}=\angle{IAE}=\angle{IAL}$, $LIAT$ is a cyclic quadrilateral Since $\angle {ETA}=\angle{ALI}=\angle{AKI}$ it is $LI||KE$. Also, since $KE||BC$, $LI||BC$. Let $TL\cap BC=S$. It is clear that "$DT$ bisects $[IF]$ $\Longleftrightarrow$ $SILF$ is a parallelogram". We have to prove that $\frac{|DI|}{|DA|}=\frac{|DS|}{|DL|}$. Let $AD\cap BC=R$. We know that $|DI|=|BD|$ (the rule that $D$ is the center of $(BIC)$.) In last equation, the expression on the right is equal to $\frac{|DR|}{|DI|}$ from $CR||IL$. If we substitute, the expression we have to prove is $\frac{|DR|}{|DI|}=\frac{|DI|}{|DA|}$. Since $\angle{CBD}=\angle{DAC}=\angle{DAB}$, $\triangle {DRB}\sim \triangle {DBA}$. $\frac{|BD|}{|DA|}=\frac{|DR|}{|BD|}$. With the equation $|BD|=|DI|$ we just said, the expression goes into $\frac{|DR|}{|DI|}=\frac{|DI|}{|DA|}$. We're done.

And one solution with 0 bash and 0 smilarities. $K = EI \cap \Gamma$, $L = AI \cap BC$, $F' = AF \cap \Gamma$, $X = KD \cap AF'$, Let $I_A$ be the $A$-excenter. We just have to show that $D, G, X$ are colinear. Now by pascal's theorem on $AF'EKDD$ we have $IP \parallel BC$ By menelaus on $AIF$ and $D-G-X$ we have to show that $\frac{AD}{ID}\frac{IG}{FG}\frac{FX}{AX}=1\Longleftrightarrow\frac{AD}{ID}=\frac{AX}{FX}$ But from $IP \parallel BC$ we have $\frac{FX}{AX}=\frac{LI}{IA}$. Let $a=AI, b=IL, c=LD$. We have to show that $AI.ID=AD.IL$ or $a(b+c)=(a+b+c)b$ or $ac=bc+b^2$. By power of Point in $ABDC$ and $CIBI_A$ we have $AL.LD=CL.LB=IL.LI_A$ or $(a+b)c=b(2c+b)$ or $ac=bc+b^2$ and we are happy .

finally broke my geo-failure streak