Graph: File:ISL2012_G6_daniel73.PNG [Amir Hossein: File attached.] Solution:

If ABC Isoscel then the second common point of circles with diameter $CO$ and $ABO$ lies on the median $BD$ (trivial by Reim) then take K' on BD such that I inercenter, O lies on $ABK'$ and OK' perp to K'I =>I-K'-C collinear

Let $O$ be the circumcentre of triangle $ABC$, let $D',M$ be the midpoints of $AB,BC$ respectively, let $E'$ be the second intersection of $AM$ with $(DBC)$. Now $\angle AFD=180^{\circ}-\angle AFB=180^{\circ}-\angle AEB=\angle BEM=\frac{1}{2}\angle BEC$ $=\frac{1}{2}\angle BDC=\frac{1}{2}(\angle AFD+\angle DAF)$ and so $\angle AFD=\angle DAF$ hence $DF=DA=DC$ and so $D$ lies on the the circle $\omega_1$ which has centre $D$ and radius $DA$. Clearly $M$ also lies on $\omega$ as $\angle AMC=90^{\circ}$. Let $\omega_2,\omega_3$ be $(DBC),(AEFB)$ respectively. Let $\omega_1$ and $\omega_2$ intersect at $P$. Clearly $AF$ is the radical axis of $\omega_1$ and $\omega_3$. The radical centre of $\omega_1, \omega_2$ and $\omega_3$ is the point of concurrency of the three radical axes $CP,BE$ and $AF$. But $AF$ and $BE$ intersect at $I$ and so $I$ lies on $CP$ (and hence so does $K$). Let $MD$ meet $\omega_2$ at $G$. Let $T$ be the point on the larger arc $BC$ of $\omega_2$ such that $MT=MC$. Then $TD$ is the perpendicular bisector of $MC$ and so $\angle DTM=\angle DTX=\angle DTP$ (as $DP=DT$) and so $T$ lies on $PM$. Now $TG=TC\iff\angle TDG=\angle TDC\iff \angle TDM=\angle TDC$ which is true. So $TM=TC=TG$. Thus since $BM=BG$ (note that $\triangle BMG\sim\triangle DCM$), we have that $BT$ is the perpendicular bisector of $MG$ thus $\angle BTG=\angle BTM\implies BP=BG$, so $BP=BM=BG$ which means that $BD$ is the perpendicular bisector of $PM$. The next paragraph is not strictly part of the solution, but I found it quite interesting: we prove that $F$ is the incentre of triangle $PBC$ (and who knows, it might help someone else come up with a solution). Let $\angle FCM=\theta$. Then $\angle FDM=2\theta\implies\angle BDG=2\theta\implies BCG=2\theta$. Note that $F$ lies on the perpendicular bisector $BD$ of $PM$ and so $FP=FM$ hence $\angle PDM=4\theta\implies\angle PDG=4\theta\implies\angle PCG=4\theta$. Thus $4\theta=\angle PCG=\angle BCG+\angle FCM+\angle PCF=3\theta +\angle PCF$ hence $\angle PCF=\theta$. Then $\angle PCF=\angle BCF$ and thus $CF$ is the angle bisector of $\angle PCB$. Also since $DB=DP$, we have that $\angle PBD=\angle CBD$ i.e. $BF$ bisects $\angle PBC$. Thus $F$ is the incentre of $PBC$. Then $\angle FPB=\angle FPC$ and so $E'$ lies on $PF$ as $PE'$ also bisects $\angle BPC$ due to the fact that $E'B=E'C$. Back to the solution. Since $BG=BP$ we have that $\angle BDM=\angle PCB\implies \angle KDM=\angle KCM$ so $K$ lies on $(CDM)$. Seeing as $\angle OMC=\angle ODC=90^{\circ}$ we have that $O$ lies on this circle too. Then $\angle OKD=\angle OCD=90^{\circ}-B=\angle MAB=\angle OAB$ so $\angle OKB+\angle OAB=180^{\circ}$. Hence $A,B,O,K$ are concyclic. Note that $\angle OKC=\angle OMC=90^{\circ}$ so $OK\perp CP$. Let $O'$ be the point diametrically opposite $O$ on $(AOKB)$. Now $\angle OKO'=90^{\circ}=\angle OKP$ so $O'$ lies on $KP$ i.e. it lies on $KI$. But since $OA=OB$ clearly $O'A=OB'$, so $O'K$ bisects $\angle AKB$ i.e. $KI$ is the angle bisector of $\angle AKB$. Finally, $D'$ lies on $\omega_2$ and $ED'=ED$ by symmetry, so we have that $\angle EB'D=\angle EBD$. Thus $BE$ bisects $\angle ABK$. Therefore the incentre of triangle $AKB$ is the intersection of the angle bisectors $BE$ and $KI$, which is of course $I$.

here is an approach using isogonal conjugates: reflect $I$ wrt $AE$ and get $J$; let $G$ be midpoint of $BC$ $\angle ADE = \angle CBE = \angle ECB = \angle EDB$ and $\angle BAE = \angle EAD$ thus $E$ is incenter of $\Delta ADB$ implying $\angle DBE = \angle EBA$ from the above we easily get $AD=DF$ and thus points $A,F,G,C$ lie on the circle with center $D$, thus $\angle FCG = \angle FAE = \angle FBE = \angle IBA = \angle ACJ$ and of course $\angle BAF = \angle JAC$ hence $F,J$ are isogonal conjugates wrt $\Delta ABC$ and we get $\angle FBA = \angle CBJ = \angle ICB$ then we get easily $\angle KCF = \angle ACJ$ easy to get $\angle ACJ = \angle EAJ$ implying that $EA$ tangents to $(ACJ)$ hence $EF^2 = EA^2 = EJ \cdot EC$ thus $EF$ tangents to $(CFJ)$ thus $\angle JFE = \angle JCF$ and so $\angle JFA = \angle DCF = \angle CFD$ so $K,J$ are isogonal conjugates wrt $\Delta FAC$ thus $\angle FAK = \angle JAD = \angle BAF$ and this implies that $I$ is incenter of $\Delta KBA$

Here is the different approach from me Claim 1 : $BI$ bisects $ABK$ Proof : Let the circumcircle of $DBC$ cuts $AB$ again at $D'$. Obviously $D'$ is the midpoint of $AB$, and hence $D'D||BC$. Take a focus at the circumcircle of $DBC$. Then $D'B=DC$, and since $E$ is on the perpendicular bisector of $BC$, then $BE=EC$. It implies that $D'E=ED$. Hence $\angle D'BE=\angle EBD$, which implies that $\angle ABI=\angle IBK$. Claim 1 proved Claim 2 : $AE=EF$ Proof : By claim 1, we get that $\angle ABE=\angle EBF$, hence looking at the circumcircle of $ABE$ we get that $AE=EF$. Claim 2 proved Claim 3 : $AF\bot FC$ Proof : Let $\angle ABI=k$ and $\angle ABC=2\alpha$. Denote by $G$ the midpoint of $BC$, hence $\angle GAC=\alpha$. We know that $\angle EBF=\angle EAF=\alpha -k$, and by claim 2 we get that $\angle ABE=\angle EBF=\alpha -k$, hence $\angle AFD=\angle BAF+\angle ABF=k+2(\alpha -k)=2\alpha -k$, which is (by simple angle chasing) same as $\angle FAD$. Hence $\Delta AFD$ is isosceles and so as $\Delta DFC$ (since $DA=DC$). Again by simple angle chasing we get that $AF\bot FC$. Claim 3 proved. Claim 4 : $AIHC$ is cyclic, where $H$ is the intersection point between $AG$ and $BF$ Proof : We can easily see that $\angle IEH=\angle HEC$. Next, note that $\angle IAE=\alpha -k=\angle ABI$, hence $EA$ is tangent to the circumcircle of $AEB$, and then $EA^2=EI\times EB$. But $\angle AFH=90$ and $AE=EF$, then $E$ is the midpoint of $AH$, it implies that $EA=EH$. Moreover, $EB=EC$, then $EH^2=EI\times EC$. Then $\Delta EIH$ is similar to $\Delta EHC$. Hence $\angle EHI=\angle ECH$, and finally $\angle IAC+\angle IHC=\angle HEC +(\angle HCE+\angle EHC)=180$, hence $AIHC$ is cyclic, claim 4 proved. Claim 5 : $CF$ bisects $\angle BCI$ Proof : By claim 5 we get that $\angle FCI=\angle IAH=\alpha -k$. By simple angle chasing we get that $\angle FCA=90-2\alpha +k$, and then $\angle FCB=(90-\alpha )-\angle FCA=\alpha -k$. Claim 5 proved. Next, we get that $\angle ABF=\angle BCI\rightarrow \angle KBC=\angle KCD$. Then $CD^2=DK\times DB$. Since $DC=DA$, then $AD^2=DK\times DB$, then $\angle AKD=\angle BAD=2\alpha$. Then we proceed to the final claim Claim 6 : $KI$ bisects $AKB$ Proof : Note that $\angle AIC=\angle AHC=90+\alpha -k$, then $\angle FIK= 90-\alpha +k$. From $\Delta BIF$, we get that $\angle IFK=\angle IBF+\angle BIF=2\alpha -k$. It implies that $\angle FKI=90-\alpha$, and then $\angle IKA=180-2\alpha -(90-\alpha )=90-\alpha$. Claim 6 proved. By claim 1 and claim 6, we get that $I$ is the incenter of $KAB$. Our proof is complete now.

I think construction part of my solution is a new one... Let $M$ be the midpoint of $BC$. Fact:1) $AD=DF=\frac{1}{2}AC$ Proof: Using cyclic property $\angle DFA=\angle MEB =\frac{1}{2}\angle BEC=\frac{1}{2}\angle BDC$ ; It implies, $\angle DFA=\angle DAF$;So, $AD=DF$ Fact:2) $BI$ bisects $\angle ABK$ Proof: By symmetry $\angle EBA=\angle ECA = \angle EBD$. Proved. Fact:3) $DA$ touches circle $BKA$ Proof: Extend $FD$ to $L$ such that $DL=DA=DF=DC$;So, $AFCL$ is rectangle. Using Fact 2)$\frac{BF}{AB}=\frac{FI}{IA}\rightarrow \frac{BF+AB}{AB}=\frac{FI+AI}{AI} \rightarrow \frac{BL}{AC}=\frac{CL}{AI}$ But as $\angle CAI = \angle CLB$; $\triangle AIC ~ \triangle LCB$ and it implies $\angle DCK = \angle KBC$; So, $DC$ touches circle $BKC$. So, $DC^2=DK.DB=DA^2$. So our claim is true. Fact:4) $AI$ bisect $\angle BAK$ Proof: i)$\angle AFD = \angle BAI + \angle ABF$ ii)$\angle DAF=\angle KAI + \angle KAD$ Fact :1 implies, $\angle AFD =\angle DAF$ Fact :3 implies, $\angle ABF = \angle KAD$ It implies, $\angle BAI =\angle KAI$ and thus our claim is true. Finally, Fact 2) and Fact 4) implies $I$ is the incenter of $\triangle KAB$ (Proved )

First note $\angle {EBA}=\angle {ECA}=\angle {DBE}$. So $BI$ is angle bisector of $\angle {DBA}$. Now suppose extended $CK$ meets $AB$ at $k'$. Suppose $\angle {EBA}=\theta,\angle {KCA}=\alpha,\angle {KAC}=x$. Now it's enough to show $\frac {AK}{BK}=\frac {AK'}{BK'}$. Now observe $\frac {AK}{BK}=\frac {Sin \alpha.Sin(B-2\theta)}{Sin(C-\alpha).Sin x}$. Also $\frac {AK'}{BK'}=\frac {Sin \alpha .Sin B}{Sin(C-\alpha).Sin A}$. Also $\frac {Sin \alpha}{Sin (C-\alpha)}=\frac {Sin\theta Cos(B-\theta)}{Cos(B+\theta)Sin(B-\theta)}$. Again $\frac {AB}{BC}=\frac {Sin x.Sin(C-\alpha)}{Sin(A-x).Sin \alpha}$. Now note $Sin 2\theta.Sin B=SinA.Sin(B-2\theta)\implies \frac {Sin B}{Sin A}=\frac {Sin(B-\theta).Cos\theta}{Sin(B+\theta).Cos (B-\theta)}$ and hence $\frac {Sin x}{Sin(A-x)}=\frac {Sin 2\theta}{Sin (2B+2\theta)}$. Thus we get $x=2\theta$.Now clealry we've $\frac {SinB}{Sin A}=\frac {Sin(B-2\theta)}{Sin x}$.So finally we obtain $\frac {AK}{BK}=\frac{AK'}{BK'}$ and that implies $KK'$ is angle bisector of $\angle {AKB}$.Thus finally done.

Note that $CE=BE$. Since $BFEA$ and $CDEB$ are cyclic, we have $\angle FBE=\angle ECA$ and $\angle AFB=\angle BEA=\angle CEA$, since $AE$ bisects $\angle BEC$. This means that $\triangle FIB\sim\triangle EAC$. Let $X$ be the midpoint of $BC$. Lemma 1: $\angle DBE=\angle EBA$ Since $AB=AC$ and $BE=CE$, $\angle EBA=\angle ACE=\angle DBE$ as $CDEB$ is cyclic. $\blacksquare$ \[\angle DFA=\angle BEX=\angle CEX=\angle CAX+\angle ECA=\angle CAX+(\angle FAE=\angle EAC)\] Thus $DA=DF=DC$. Then as $DX=DA=DC$, $AFXC$ is cyclic. Lemma 2: $CA$ is tangent to the circle $AKB$. We will first show that $CD$ is tangent to the circle $CKB$. Let $CK$ intersect the circle $CDEB$ at $M$. Then by PoP $MI\cdot IC=IE\cdot EB=IF\cdot IA$, thus $FCAM$ is cyclic. Since $AFXC$ is cylic, $AMFXC$ is cyclic with center $D$, and $\angle DMC=\angle DCM$. However, $\angle DMC=\angle DBC$ since $CDEBM$ is cylic, thus $\angle DBC=\angle DMC=\angle DCM$. Then $\angle DCK=\angle DBC$, and $CD$ is tangent to the circle $BKC$. By PoP $CD^2=DK*DB$, and as $DC=DA$, $DA^2=DK*DB$ implying that $CA$ is tangent to the circle $AKB$. $\blacksquare$ By lemma 2 $\angle DAK=\angle DBA$. As $\angle EAI+\angle AFE=\angle FBA$ as $ABFE$ is cyclic, $\angle EAI+\angle AFE=\angle DAK$, implying that $\angle DAE=\angle AFE+\angle FAE+\angle EAK=\angle IAK+\angle EFA$. But $\angle AFE=\angle EAF$, implying that $\angle BAX=\angle BAE+\angle EFA$. But as $\angle BAX=\angle CAX$, $\angle BAE=\angle IAK$. Thus with lemma 1 and the above, we have that $I$ is the incenter of $\triangle KAB$.

my method $S=AF/cap BC$ $M$ is midpoint of $BC$ $N=CI/cap BD$ first notice $F$ is on $k(D,DA)$ and $BE$ bisects $\angle DBA$ so we need $\angle CIA=90+x$ where $x=\angle CAF$. or only $\angle SCI=2x$. now first menelaus on $FDA$ and using $BI/IA=BF/BA$ and $AC=2*DF$ we get $FN=AC*BF/(BA+2*BF)$. now we intersect circle $CMD$ with $BD$ at $N'$ and it's enough to prove $N'=N$ cause then $\angle NCB=\angle MDB=\angle DBA=2x$ now notice $MNF\sim MFS\sim SCA$ and with some ratio chase we get $FN'=SF*AC/SA$ so we only need $SF/SA=BF/(BA+2*BF)$ flip this and subtract by $1$ to get $AF/FS=(BA+BF)/BF$ but menelaus on $CSA$ gives $FA/FS=CB/BS$ so we need $CB/BS=(BA+BF)/BF$ but subtracting by $1$ we only need $BS/CS=BF/BA$ which is true by menelaus on $CDB$.

I solved this by using many Menelaus and Ceva Theorems. It was a big ratio chase, followed by some expanding. I will post my solution later when I have time.

It's easy to deduct that $BE$ bisects $\angle ABD$, thus I is the incenter of triangle $KAB$ $\iff$ $\angle AIC=\angle ABE+90$. Since $AE$ bisects $\angle BAD$, thus $E$ is the incenter of triangle $ABD$, which means that it suffices to prove $\angle AIC=\angle ABE+90=\angle AED$. Let $X$ be on $DE$ be on such that $\angle DAX=\angle FAE$. Since $A,B,F,E$ are concyclic, thus $\angle DAX=\angle FAE=\angle EBF=\angle ACE$, which means that $CE \parallel AX$. Since $AD=DC$, thus $AXCE$ is a parallelogram, which means that $\angle AXC=\angle AEC=\angle AEB=\angle AEI$ $\Rightarrow$ triangles $AXC$ and $AEI$ are similar $\Rightarrow$ triangles $AEX$ and $AIC$ are also similar by spiral similarity, which means that $\angle AIC=\angle AED$ and that's our desired result.

Easy problem. BI is bisector of KAB obviously. Now if we prove KIA=90+(B/2) we're done. By angle chasing that reduces to proving ICA=DBC. Take the circumcircle of AFC which intersects the circumcircle of DBC in I'. Because of radical center theorem CII' are collinear, so proving I'CA=DBC finished problem. We will now prove CFA=90. Call BDC=$\alpha$, DBA=$\beta$, so if F' is intersection of BD with the circle center D radius DA, it suffices to prove BFEA is cyclic, which reduces to $\frac{\beta}{2}=BAE=BFC=C-90+\frac{\alpha}{2}$, which reduces to $90-C=\frac{\alpha-\beta}{2}$, or $A+\beta=\alpha$, which is true. So CFA=90 so DC=DF=DA=DI' so I'CA=DI'C=DBC and we're done. Note that CI is the simedian of ABC.

Hybrid solution! This is actually quite easy with barycentrics in the sense that you can that you can substitute any missed synthetic observations with computations and still come out unscathed. The solution below was as optimized as I could get (the nontrivial step of the proof is about 20 seconds of determinants.) [asy][asy] size(9cm); pointpen = black; pair B = Drawing("B", (-3,0), dir(225)); pair C = Drawing("C", (3,0), dir(-45)); pair A = Drawing("A", (0,7), dir(90)); pair D = Drawing("D", midpoint(A--C), dir(A+C)); pair D1 = Drawing("D'", midpoint(A--B), dir(A+B)); pair E = Drawing("E", incenter(A,D,B), dir(90)); pair F = Drawing("F", reflect(D,E)*A, dir(-70)); pair I = Drawing("I", IP(B--E,A--F), dir(135)); pair K = Drawing("K", IP(C--I,B--D), dir(270)); draw(A--D--B--cycle, black+1); draw(circumcircle(D,B,C), dashed); draw(D1--D--C--B, dotted); draw(B--E); draw(A--F); draw(C--I); draw(A--K, dashed); [/asy][/asy] Let $D'$ be the midpoint of $\overline{AB}$. Evidently the points $B$, $D'$, $D$, $E$, $C$ are concyclic. By symmetry, $DE=D'E$, and hence $\overline{BE}$ is a bisector of $\angle D'BD$. It follows that $E$ is the incenter of triangle $ABD$. Since the center of $(AEB)$ lies on ray $DE$ by a well-known fact (which I like to call Fact 5), it follows that the reflection of $A$ over $\overline{ED}$ lies on $(AEB)$, and hence is $F$. We now claim that $DK \cdot DB = DA^2$. The proof is by barycentric coordinates on $\triangle ABD$. Set $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$ and let $a = BD$, $b = AD$ and $c = AB = 2b$. The observations above imply that $F = (0 : b : b-a)$ and $E = (a : b : c)$. This implies $I = \left( a(a-b) : bc : c(a-b) \right) = \left( a(a-b) : 2b^2 : 2b(a-b) \right)$. Finally, $C = (-1,0,2)$. Hence if $K = (0:y:z)$ then we have \[ 0 = \left\lvert \begin{array}{ccc} 0 & y & z \\ -1 & 0 & 2 \\ a(a-b) & 2b^2 & 2b(a-b) \end{array} \right\rvert = \left\lvert \begin{array}{ccc} 0 & y & z \\ -1 & 0 & 2 \\ 0 & 2b^2 & 2(a^2-b^2) \end{array} \right\rvert \] so $y : z = b^2 : (a^2-b^2)$, so $K = \left( 0, \tfrac{b^2}{a^2}, 1-\tfrac{b^2}{a^2} \right)$. It follows immediately that $DK = \tfrac{b^2}{a}$ as desired. Now remark that $DK \cdot DB = DA^2 \implies \triangle DAK \sim \triangle DBA \implies \angle FAD = \angle B$. So $\angle BAK = \angle A -\angle B$. On the other hand, $\angle EAD = %Error. "half" is a bad command. \angle A$ and $\angle FAE = \angle FBE = %Error. "half" is a bad command. \angle B$ implies $\angle BAF = %Error. "half" is a bad command. (\angle A- \angle B)$, so we're done.

At first we prove the following lemma. Lemma: $ABC$ be a triangle where $AB=AC$.Let $P$ be a brocard point of $\bigtriangleup ABC$ such that $\angle ABP=\angle BAP=\angle CAP$.Then $BP$ is a median. Proof: Let $BP$ meets $AC$ at $Q$.After some angle chasing we get $\angle APQ=\angle A,\angle CPQ=\angle B$. Again in $\bigtriangleup ACP$ and $\bigtriangleup BCP$ $\angle CAP=\angle BCP,\angle ACP=\angle C-\angle BCP=\angle B-\angle ABP=\angle CBP$ So $\bigtriangleup ACP \sim \bigtriangleup BCP$ and $\dfrac{AP}{CP}=\dfrac{AC}{BC}$ So $\dfrac{AQ}{BQ}=\dfrac{AP.sin APQ}{CP.sin CPQ}=\dfrac{AC.sin A}{BC.sin B}=\dfrac{AC.BC}{BC.AC}=1$ SO $Q$ is the midpoint of $AC$.(proved) Main proof: By our lemma,$P$ lies on $BD$.In $\bigtriangleup AEC$ and $\bigtriangleup ABE$ $AC=BC,AE=BE$ and $\angle CAE=\angle BAE$. So $\bigtriangleup ACE \cong \bigtriangleup ABE$ $\angle ABI=\angle ABE=\angle ACE=\angle DCE=\angle DBE=\angle \angle PBI$.So $BI$ bisects $\angle ABP$ Again $\angle PAF=\angle AFP-\angle ABP=\angle AEB-\angle ADP-\angle DAP$ $=180^{\circ}-\dfrac{\angle BEC}{2}-\angle ADP-\angle DAP$ $=90^{\circ}+\dfrac{180^{\circ}-\angle BDC}{2}-\angle ADP-\angle DAP=90^{\circ}+\dfrac{\angle ADB}{2}-\angle ADP-\angle DAP$ $=\dfrac{\angle DAP+\angle APD}{2}-\angle DAP=\dfrac{\angle APD-\angle DAP}{2}=\dfrac{\angle BAP}{2}$ So $\angle PAF=\angle BAF$. So $I$ is the incenter of $\bigtriangleup ABP$ $\angle API=\angle BPI$ $\angle API=\angle API=\dfrac{\angle APB}{2}=90^{\circ}-\dfrac{\angle A}{2}=\angle C$ Again $\angle APC=180^{\circ}-\angle C$ $\angle APC+\angle API=\angle C+180-\angle C=180^{\circ}$ So $C,P,I$ collinear So $P=CI \cap BD$ Again $K=P=CI \cap BD$ So $P\equiv K$ This implies $I$ is the incenter of $\bigtriangleup KAB$.

Nice Problem! My solution:$BI$ is bisector of $\angle KAB$ obviously... so we have $\overline{AE}=\overline{EF} \Leftrightarrow \angle EAF=\angle EFA$. So \[\angle AJD=\angle AJ'D+\angle BAF=\frac {\angle A}{2}+\angle B=\frac {\pi}{2}\] so we have $\overline{AD}=\overline{DF}=\overline{DC}$. It means that \[\angle AFC=\frac {\pi}{2} \Leftrightarrow \angle FCB=\angle ABI \] Q.E.D.

Let $\angle{ABE}=\theta$.Clearly $\triangle{ABE} \cong \triangle{ACE} \Rightarrow \angle{ECA}=\angle{EBD}=\theta \Rightarrow IB$ bisects $\angle{ABK}$.Also it follows from the fact that $\angle{FAE}=\theta$.From the cyclic $BCDE$ we get $\angle{EDA}=\angle{EBC}=B-\theta \Rightarrow ED \perp AF$.This along with $AE=EF \Rightarrow DF=AD=DC=\frac{b}{2} \Rightarrow AF \perp CF$.Let $CF \cap AE=X$.One can easily see that $AE=EX$ and $EB=EC$.Now $\angle{IAE}=\angle{IBA}=\theta \Rightarrow AE$ is tangent to the circumcircle of $\triangle{AIB} \Rightarrow AE^2=EX^2=EI \times EB=EI \times EC \Rightarrow \triangle{EIX} \sim \triangle{EXC} \Rightarrow \angle{IAC}+\angle{IXC}=\angle{IXA}+\angle{AXC}+\angle{IAC}=\angle{IKE}+\angle{XIE}+\angle{IEX}=180^{\circ} \Rightarrow AIXC$ is cyclic. (where we have used the fact that $\angle{ECD}=\theta$)Note that $E$ is the incenter of $\triangle{ABD}$.Finally $\angle{AIK}=\angle{AIC}=\angle{AXC}=\angle{AED}=90^{\circ}+\frac{\angle{ABD}}{2}$ and the result follows.

Very nice problem. I think the following approach hasn't come up yet, even though it bears some similarities to some of the ones posted above. First, the "easy" angle bissector: let $D'$ be the midpoint of $AB$. Because of trivial symmetry, the circle through $B$, $C$, $D$ (which we'll call $\omega$) passes through $D'$, and $E$ is the midpoint of arc $DD'$ of that circle. Thus, $D'\widehat{B}E=E\widehat{B}D$, or equivalently $A\widehat{B}I=I\widehat{B}K$. Therefore, it suffices to prove that $B\widehat{K}I=I\widehat{K}A$. First, we claim that $F$ lies on the circle with diameter $AC$. This follows from an easy angle-chase: we have $A\widehat{F}D=\pi-B\widehat{F}A=\pi-B\widehat{E}A=\frac{1}{2}\cdot C\widehat{E}B=\frac{\pi -F\widehat{D}A}{2}$. Therefore, $DA=DF$, and since $D$ is the midpoint of $AC$, $F$ lies on the circle with diameter $AC$, establishing our claim. Let $\omega_1$ be the circle with diameter $AC$. Now, since \[IE\cdot IB=IF\cdot IA\] $I$ has the same power with respect to $\omega$ and $\omega_1$, and therefore it lies on the radical axis of the two circles. Since $C$ trivially lies on the radical axis as well, $CI$ is the radical axis of $\omega$ and $\omega_1$, and $K$ lies on the radical axis. Let $F'$ be the reflexion of $F$ over $D$, which lies on $\omega_1$. Then, we have \[KF'\cdot KF=KD\cdot KB\] or, equivalently, \[(DF+DK)\cdot(DF-DK)=DK(DB-DK)\] \[DF^2=DK\cdot DB\] \[DA^2=DK\cdot DB\] which implies $\frac{DK}{DA}=\frac{DA}{DB}$, and this implies from sas that triangles $DKA$ and $DAB$ are similar. Thus, $\frac{AK}{DK}=\frac{AB}{AD}=\frac{AC}{AD}=2$. Finally, let $M$ be the midpoint of $AK$; we then have $KM=KD$. Notice that $DM$ is parallel to $CK$, and this implies that $CK$ is the external angle bissector of $\angle{MKD}$, or, equivalently, the internal angle bissector of $\angle{BKA}$. This establishes the problem statement.

A nice problem.First,let $D'$ be the midpoint of $AB$ and let $L$ be the second intersection point of $BF$ with the circumcircle of $AFC$.Now,it is easy to prove that $BE$ is the angle bisector of angle $DBA$(just use that $ED'=ED$).Now,using this proven fact we note that $AE=EF$ and now we get that triangles $AD'E$ and $FED$ are congruent,from which we have $AD'=FD=AD=CD$ from which we have that angle $AFC$ is right.Now,from an easy angle chase we conclude that we need $DK*DB=DA^2$,but this is easy to prove cause $CK$ is the radical axes of $AFC$ and $ABD$,so we have $KF*KL=KB*KD$($FCLA$ is a rectangle)or $AD^2-DK^2=DK*DB-KD^2$,so we are done.

Firstly notice $BI$ is bisector so it is enough to prove $\angle AIC = 90+(\angle ABD/2)$. But using a trigbash it can be shown fairly simply that $\angle AFC=90$, and let $X$ be the intersection of the circle with diameter $AC$ and the circumcircle of $BDC$, and it seen that $I$ is the radical center of the three circles. Now using $\angle DXC = \angle DBC$ one can show easily that $XF=FM$ (the arcs), where $M$ is the midpoint of $BC$ which also lies on the circle with diameter $AC$, and now it is just an angle chase.

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Notice that by spiral similarity lemma on $(BGDC)$ and $(AFEB)$ we have $$\triangle EGD\sim\triangle EAF$$Therefore $\frac{EA}{EF}=\frac{EG}{ED}=1$, hence it suffices to show $\angle BAK=2\angle BAI$. Notice that $\angle EFD=\angle BAE=\angle DAE$ hence $DA=DF=DC$, hence $\angle AFC=90^{\circ}=\angle EMC$, meanwhile $$\angle MEC=\angle BEM=\angle AFD=\angle FAC$$Hence $\triangle CEM\sim\triangle CAF$. CLAIM. $CI$ is the $I$-symmedian of $\triangle ACB$. Proof. Let $\angle BAD=A, \angle ABD=B$ and $\angle ADB=D$, then by Ceva's theorem and easy angle chasing, \begin{align*} \frac{\sin\angle BCI}{\sin\angle IAC}\cdot\frac{\sin\angle IBC}{\sin\angle ABI}&=\frac{\sin\frac{\angle DBC}{2}}{\sin\angle MEC}\cdot\frac{\sin\angle ECM}{\sin\frac{\angle ABD}{2}}\\ &=\frac{\sin\frac{A-B}{2}}{\cos\frac{D}{2}}\cdot\frac{\sin\frac{D}{2}}{\sin\frac{B}{2}}\\ &=\frac{\sin A+\sin B}{\cos\frac{A}{2}+\sin\frac{B-D}{2}}\\ &=2\sin\frac{A}{2} (\text{this uses the fact that }2\sin B=\sin D)\\ &=\frac{\sin\angle ACG}{\sin\angle GBC} \end{align*}This proves the CLAIM. $\blacksquare$ Now \begin{align*} \angle BAI&=\angle BAC-\angle FAC\\ &=180^{\circ}-2\angle ACB-90^{\circ}+\angle ECM\\ &=90^{\circ}-2\angle ACB+\angle ECM\\ &=\angle EAC-\angle ACE \end{align*}Now $\angle ACK=\angle BCG=\angle DBC$, hence $K$ is the $B-$ humpty point of $\triangle ABC$. Hence $$\angle KAC=\angle ABD=\angle GCA$$This implies $$\angle BAK=\angle GAC-\angle GCA=2(\angle EAC-\angle ACE)=2\angle BAI$$as desired.

Let $J=\bigtriangleup{ABC}\cap{(BDC)}$ Since E is the intersection of the bisector of $\angle{BAC}$ with $(BDC)$ , and $AB=AC$ , and $M$ is the midpoint of $AC$ hence $E$ is the midpoint of the arc $\overarc{JD}$ , thus $\angle{ABI}=\angle{JBE}=\angle{EBD}=\angle{IBK}$.$(1)$ Now $\angle{EFD}=\pi-\angle{EFB}=\angle{EAB}=\angle{EAD}$ , and $\angle{EFA}=\angle{EAF}$ , thus $\angle{DAF}=\angle{DFA}$ , since $D$ is the midpoint of $AC$ , hence $DA=DC=DF$ , thus points $A , F , C$ lie on the circle with diameter $AC$ , wich implies that $\angle{AFC}=\frac{\pi}{2}$. And let $M=AE\cap{BC}$ , since $\triangle{ABC}$ is isoceles at $A$ , thus $\angle{AMC}=\frac{\pi}{2}$ , thus $AFMC$ is cyclic . Now let $N=IC\cap{(BDC)}$ , thus $IC\times{IN}=IE\times{IB}=IA\times{IF}$ , hence $N$ lies on $(AFC)$ , it implies that $DC=DN$. $\angle{KCD}=\angle{NCD}=\frac{\overarc{ND}}{2}=\frac{\overarc{CD}}{2}=\angle{DBC}=\angle{KBC}$. $(\alpha)$ Let $K'$ be the reflection of K throught D , thus $AK'CK$ is a parallelogram , hence $\angle{DBC}=\angle{KCD}=\angle{K'AC}$ , wich implies that $K'$ lies on $(ABC)$. Let $L=CF\cap{(BDC)}$ , $\angle{FDC}=\angle{BDC}=\angle{BEC}$ , and since $\triangle{FDC}$ and $\triangle{BEC}$ are isoceles at $D$ and $E$ respectively , thus $\angle{DCF}=\frac{\overarc{DL}}{2}=\angle{ECB}=\frac{\overarc{EB}}{2}$ , and $\angle{DFC}=\frac{\overarc{CD}+\overarc{LB}}{2}=\angle{DCF}=\frac{\overarc{LD}}{2}=\frac{\overarc{LN}+\overarc{ND}}{2}=\frac{\overarc{LN}+\overarc{CD}}{2}$ , it implies that $\overarc{LB}=\overarc{LN}$ , and since $\overarc{ED}=\overarc{LB}$ , thus $\overarc{JD}=\overarc{NB}$. On the other hand , $\angle{KAC}=\angle{ACK'}=\angle{ABK'}=\frac{\overarc{JD}}{2}=\frac{\overarc{NB}}{2}=\angle{KCB}$ . $(\beta)$ Thus from $(\alpha)$ and $(\beta)$ we deduce that $\triangle{AKC}$ and $\triangle{CKB}$ are similar . Now $\angle{IKA}=\pi-\angle{AKC}=\pi-\angle{BKC}=\angle{IKB}$.$(2)$ Hence from $(1)$ and $(2)$ $I$ is the incenter of $\triangle{AKB}$

I don't like this problem, but anyway, I'll post my solution. Let $D'$ be the midpoint of $AB$ and $G$ be the centriod of $\triangle ABC$. First, we have $\angle ABE = \angle ECD = \angle EBD$, hence $BE$ is the angle bisector of $\angle ABK$. Now we're enough to show that $\angle AIK = 90^\circ + \frac{1}{2} \angle ABK$. Let $M$ be the midpoint of $BC$. Notice that $\angle AFD = \angle MEB = \angle CEM = \angle ECA + \angle CAE + \angle EBD + \angle CAE = \angle EAF + \angle CAE = \angle CAF$. Hence $AD = DF$, we have $F\in \odot (AC)$. And since $\angle FDC = \angle BDC$ and $DC = DF, EC = EB$, $\triangle CDF \stackrel{+}{\sim} \triangle CEB$, which shows that $CE, CF$ are isogonal WRT $\angle ACB$. By Desargues Involution $CG, CI$ are isogonal WRT $\angle ACB$. Let $CI$ intersect $\odot (BCD)$ at $T (\neq C)$. Since $I$ is the power center of three circle $\odot (BCD), \odot (ABE), \odot (AC)$, hence $T \in \odot (AC)$. So $\angle AIK - 90^\circ = \angle TAI = \angle ICF = \angle DCE = \angle DBE$, we're done.

Mostly bary, with minor synthetic observations. Let $D_1 = D$, and $D_2$ the midpoint of $AB$. Observe that $D_2\in (BCD_1)$, and $E$ is the arc midpoint of $D_1D_2$, so $\angle D_2BE = \angle EBD_1$. Thus, it suffices to show that $KI$ bisects $\angle BKA$, or that $AK$ is the external angle bisector of $\angle AKD_1$. If we define $N$ as the point in segment $AD_1$ such that $AN = 2ND_1$, then showing that $\angle NKC = 90$ is enough by Apollonius circles. Now, we bash it out. Let $a = AB = AC, b = BC$. We have $A = (1,0,0), B = (0,1,0), C = (0,0,1), D_1 = (1:0:1). N = (2:0:1)$. Circle $(BCD_1)$ has the equation \[(BCD_1) = -b^2yz - a^2xz - a^2xy + \frac{a^2}{2}x(x+y+z) = 0\]If we let $E = (1 : x : x)$, then plugging this in, we get \[-b^2x^2 - 2a^2x + \left(\frac{a^2}{2}\right)(2x+1) = 0 \Rightarrow x = \frac{a\sqrt{a^2 + 2b^2}-a^2}{2b^2} \]by solving for $x$ using the quadratic formula. Then, if $F = (1 : y : 1)$, we have $I = (1 : xy : x), K = (1 : xy : 1)$. Furthermore, the equation for $(ABE)$ has the form \[-b^2yz - a^2xz - a^2xy + (vz)(x+y+z) = 0\]Plugging $E = (1 : x : x)$ into this, and using the fact that $-b^2x^2 - 2a^2x = -\left(\frac{a^2}{2}\right)(2x+1)$, we have \[vx(2x + 1) = \frac{a^2}{2}(2x+1) \Rightarrow u = \frac{a^2}{2x} = \frac{a}{\frac{a\sqrt{a^2 + 2b^2} - a}{b^2}} = \frac{ab^2}{\sqrt{a^2 + 2b^2} - a}\cdot \frac{\sqrt{a^2 + 2b^2} + a}{\sqrt{a^2 + 2b^2} + a} = \frac{a\sqrt{a^2 + 2b^2} + a^2}{2}\]Now, plugging in $F= (1 : y : y)$, we have \[-b^2y^2 - a^2 - a^2y + u(y+2) = 0\Rightarrow y = \frac{a^2 - 2u}{u - a^2 - b^2}\]\[= \frac{2a\sqrt{a^2 + 2b^2}}{a^2 + 2b^2 - a\sqrt{a^2 + 2b^2}} = \frac{2a}{\sqrt{a^2 + 2b^2 - a}}\]Finally, we have \[xy = \frac{2a}{\sqrt{a^2 + 2b^2} - a}\cdot \frac{a(\sqrt{a^2 + 2b^2 - a})}{2b^2} = \frac{a^2(\sqrt{a^2 + 2b^2} + a)(\sqrt{a^2 + 2b^2} - a)}{2b^4} = \frac{a^22b^2}{2b^4} = \frac{a^2}{b^2}\]Therefore, $K = (b^2 : a^2 : b^2)$, and \[\overrightarrow{KM} = (b^2 : a^2 : -b^2 - a^2), \overrightarrow{KN} = (-b^2 - 2a^2 : 3a^2 : b^2 - a^2)\]Now, using the perpendicularity criterion, we need to show the following is 0: \[0 = b^2((b^2 - a^2)a^2 - (b^2 + a^2)(3a^2)) + a^2((b^2)(b^2 - a^2) + (b^2 + a^2)(b^2 + 2a^2)) + a^2(3a^2b^2 - a^2(b^2+2a^2)) \]\[\Longleftrightarrow 0 = b^2(-2b^2 -4a^2) + b^2(b-a^2) + (b+2a^2)(b^2 + a^2 - a^2) + 3a^2b^2\]\[\Longleftrightarrow 0 = (-2b^2 - 4a^2) + (b^2-a^2) + (b^2 + 2a^2) + 3a^2 = 0\]Thererfore, $KM\perp KN$. We conclude the problem.

Let $P = AE \cap (DBC)$ distinct from $E$; note that since $EP$ is the perpendicular bisector of $BC$, it is a diameter of $(DBC)$. Let $N = DE \cap AF$. Claim 1: $DE \perp AF$. Proof: Let $F'$ be the point on $BD$ such that $AF' \perp DE$. We prove that $F'$ lies on $(DBC)$, hence $F' = F$. $\angle EAF' = \angle EAN = 90 - \angle NEA = 90 - \angle DEP = \angle EPD = \angle EBD = \angle EBF'$. Hence $ABF'E$ cyclic. $\square$ Claim 2: $D$ circumcentre of $(AFC)$. Proof: $\angle DAF = 90 - \angle NDA = 90 - \angle EDA = 90 - \angle EBC = \angle PEB = \angle PEF + \angle FEB = \angle ABF + \angle FAB = \angle AFK = \angle AFD$. Hence $DF = DA = DC$ by $M$ midpoint. $\square$ Claim 3: $BI$ internal angle bisector of $\angle ABK$. Proof: Note that $D$ lies on the perpendicular bisector of $AF$, and $DE \perp AF$. Hence, $ED$ is the perpendicular bisector and $BE$ is the internal angle bisector of $\angle ABF$. $\square$ Let $Q = AE \cap FC$, and $M$ the midpoint of $BC$. Claim 4: It suffices to show $\frac{FQ}{AF} = \frac{IF}{FC}$. Proof: Since $\angle AFQ = \angle IFC$, this implies $\triangle FQA = \triangle FIC \Rightarrow AIQC$ cyclic. Now $\angle KCA = \angle ICA = \angle BCA - \angle BCF - \angle FCI = \angle ABC - \angle MCF - \angle QCI = \angle ABC - \angle MAF - \angle QAI = \angle ABC - \angle ABF = \angle FBC = \angle KBC$. Hence, since $K$ lies on the $B$-median, $K$ is the $B$-humpty point of $\triangle ABC$. Now $\angle DAK = \angle ABK$. $\angle IAB = \hat{A}/2 - \angle EAF = \hat{A}/2 - \angle AFE = \hat{A}/2 - \angle ABE$. $\angle KAB = \hat{A} - \angle DAK = \hat{A} - \angle ABK = \hat{A} - 2\cdot \angle ABE$. Thus $\angle KAB = 2 \cdot \angle IAB$, and $I$ is the incentre of $\triangle KAB$. $\square$ Claim 5: $\frac{FQ}{AF} = \frac{IF}{FC}$. Proof: Let $\theta = \angle ABE$. We make extensive use of the sine rule and sine in right angled triangles, as well as similarity of right angled triangles. Note that a homothety scale factor $\frac{1}{2}$ from $A$ maps $C \to D$, $Q \to E$, since $DE \perp AF \perp FC \Rightarrow DE || QC$. \begin{align*} \frac{MC}{AC} &= \sin(\hat{A}/2) \\ \Rightarrow \frac{\sin(\theta)}{\sin(\hat{A}/2)} \frac{MC}{AC} &= \sin(\theta) \\ \Rightarrow \frac{AE}{BE} \frac{MC}{AC} &= \frac{FQ}{AQ} \\ \Rightarrow FQ &= \frac{AQ \cdot AE \cdot MC}{BE \cdot AC} \\ \Rightarrow \frac{FQ}{AF} &= \frac{AQ \cdot AE \cdot MC}{AF \cdot BE \cdot AC} \\ \frac{QC}{MC} = \frac{AQ}{AF} \Rightarrow QC &= MC \cdot \frac{AQ}{AF} \\ \Rightarrow \frac{FQ}{AF} &= \frac{QC}{AC} \cdot \frac{AE}{BE} \\ &= \frac{ED}{DC} \cdot \frac{AE}{BE}. \end{align*}Now note that since $\angle ECD = \angle EBD = \angle EAF = \angle BCF$, and $F$ lies on $BD$ whereas $E$ on $(DBC)$, $E$ and $F$ are inverses under $\sqrt{BC}$-inversion at $C$. Hence $\triangle DEC \sim \triangle FBC$. \begin{align*} \frac{FQ}{AF} = \frac{BF}{FC} \cdot \frac{AE}{BE} = \frac{BF}{BE} \cdot \frac{AE}{FC}. \end{align*}By $AA$ we have $\triangle ABE \sim \triangle IBF$. Thus \begin{align*} \frac{FQ}{AF} = \frac{IF}{AE} \cdot \frac{AE}{FC} = \frac{IF}{FC}. \; \; \square \end{align*}$\blacksquare$

Firstly , since quadrilaterals $BEDC$ and $AEFB$ are cyclic , one can see that : $$\angle IBA = \angle ECD = \angle EBD=\angle IBK=\angle FAE = \alpha $$So point $I$ lies on the bisector of angle $\angle ABK$ and for proving that $I \equiv I_{\triangle ABK}$ it's enough to show that $\angle AIK=90+\angle IBA$. Let $H_{A}$ be the foot of altitude from $A$ to $BC$ , now while $\angle IAE=\angle ACE$ , we can get $\angle CAI=\angle CEH_{A}=\angle BEH_{A}=\beta$ , so since quadrilateral $BEDC$ is cyclic , we have $\angle BDC=2\angle FAD$ and $DA=DC=DF$. So $\angle AFC=90$ and quadrilateral $AFH_{A}C$ is cyclic and as the result , we have $\angle FAH_{A}=\angle FCB=\alpha$. Now while $BD$ is a median in triangle $\triangle ABC$ , one can see that : $$\frac{\sin(\angle ABD)}{\sin(\angle DBC)}=\frac{\sin(2\alpha)}{\cos(\alpha+\beta)}=\frac{\sin(\angle BAD)}{\sin(\angle BCD)}=\frac{\sin(2(\beta-\alpha))}{\cos(\beta-\alpha)}=2\sin(\beta-\alpha) \implies \frac{\cos(\alpha+\beta)}{\sin(\alpha)}=\frac{\cos(\alpha)}{\sin(\beta-\alpha)} (I)$$Now while it's enough to show that $\angle ICF=\alpha$ , by Cevian Theorem for point $F$ is triangle $\triangle BIC$ , one can see that : $$P \iff \frac{\sin(\angle ICF)}{\sin(\angle FCB)}.\frac{\sin(\angle FBC)}{\sin(\angle IBF)}.\frac{\sin(\angle BIF)}{\sin(\angle CIF)}=1 \iff \frac{\sin(\angle FBC)}{\sin(\angle IBF)}=\frac{\sin(\angle CIF)}{\sin(\angle BIF)}$$$$ \iff \frac{\cos(\alpha+\beta)}{\sin(\alpha)}=\frac{\cos(\alpha)}{\sin(\beta-\alpha)}$$Which is true by $(I)$. So we're done.

Shayan-TayefehIR wrote: Firstly , since quadrilaterals $BEDC$ and $AEFB$ are cyclic , one can see that : $$\angle IBA = \angle ECD = \angle EBD=\angle IBK=\angle FAE = \alpha $$So point $I$ lies on the bisector of angle $\angle ABK$ and for proving that $I \equiv I_{\triangle ABK}$ it's enough to show that $\angle AIK=90+\angle IBA$. Let $H_{A}$ be the foot of altitude from $A$ to $BC$ , now while $\angle IAE=\angle ACE$ , we can get $\angle CAI=\angle CEH_{A}=\angle BEH_{A}=\beta$ , so since quadrilateral $BEDC$ is cyclic , we have $\angle BDC=2\angle FAD$ and $DA=DC=DF$. So $\angle AFC=90$ and quadrilateral $AFH_{A}C$ is cyclic and as the result , we have $\angle FAH_{A}=\angle FCB=\alpha$. Now while $BD$ is a median in triangle $\triangle ABC$ , one can see that : $$\frac{\sin(\angle ABD)}{\sin(\angle DBC)}=\frac{\sin(2\alpha)}{\cos(\alpha+\beta)}=\frac{\sin(\angle BAD)}{\sin(\angle BCD)}=\frac{\sin(2(\beta-\alpha))}{\cos(\beta-\alpha)}=2\sin(\beta-\alpha) \implies \frac{\cos(\alpha+\beta)}{\sin(\alpha)}=\frac{\cos(\alpha)}{\sin(\beta-\alpha)} (I)$$Now while it's enough to show that $\angle ICF=\alpha$ , by Cevian Theorem for point $F$ is triangle $\triangle BIC$ , one can see that : $$P \iff \frac{\sin(\angle ICF)}{\sin(\angle FCB)}.\frac{\sin(\angle FBC)}{\sin(\angle IBF)}.\frac{\sin(\angle BIF)}{\sin(\angle CIF)}=1 \iff \frac{\sin(\angle FBC)}{\sin(\angle IBF)}=\frac{\sin(\angle CIF)}{\sin(\angle BIF)}$$$$ \iff \frac{\cos(\alpha+\beta)}{\sin(\alpha)}=\frac{\cos(\alpha)}{\sin(\beta-\alpha)}$$Which is true by $(I)$. So we're done. actually I found $\angle AIK=90+\angle IBA$, but I don't know why it is enough. Can you please mention how it finishes from there?

I am sorry guys, I asked them to delete this but they didn't.

adorefunctionalequation wrote: Shayan-TayefehIR wrote: Firstly , since quadrilaterals $BEDC$ and $AEFB$ are cyclic , one can see that : $$\angle IBA = \angle ECD = \angle EBD=\angle IBK=\angle FAE = \alpha $$So point $I$ lies on the bisector of angle $\angle ABK$ and for proving that $I \equiv I_{\triangle ABK}$ it's enough to show that $\angle AIK=90+\angle IBA$. Let $H_{A}$ be the foot of altitude from $A$ to $BC$ , now while $\angle IAE=\angle ACE$ , we can get $\angle CAI=\angle CEH_{A}=\angle BEH_{A}=\beta$ , so since quadrilateral $BEDC$ is cyclic , we have $\angle BDC=2\angle FAD$ and $DA=DC=DF$. So $\angle AFC=90$ and quadrilateral $AFH_{A}C$ is cyclic and as the result , we have $\angle FAH_{A}=\angle FCB=\alpha$. Now while $BD$ is a median in triangle $\triangle ABC$ , one can see that : $$\frac{\sin(\angle ABD)}{\sin(\angle DBC)}=\frac{\sin(2\alpha)}{\cos(\alpha+\beta)}=\frac{\sin(\angle BAD)}{\sin(\angle BCD)}=\frac{\sin(2(\beta-\alpha))}{\cos(\beta-\alpha)}=2\sin(\beta-\alpha) \implies \frac{\cos(\alpha+\beta)}{\sin(\alpha)}=\frac{\cos(\alpha)}{\sin(\beta-\alpha)} (I)$$Now while it's enough to show that $\angle ICF=\alpha$ , by Cevian Theorem for point $F$ is triangle $\triangle BIC$ , one can see that : $$P \iff \frac{\sin(\angle ICF)}{\sin(\angle FCB)}.\frac{\sin(\angle FBC)}{\sin(\angle IBF)}.\frac{\sin(\angle BIF)}{\sin(\angle CIF)}=1 \iff \frac{\sin(\angle FBC)}{\sin(\angle IBF)}=\frac{\sin(\angle CIF)}{\sin(\angle BIF)}$$$$ \iff \frac{\cos(\alpha+\beta)}{\sin(\alpha)}=\frac{\cos(\alpha)}{\sin(\beta-\alpha)}$$Which is true by $(I)$. So we're done. actually I found $\angle AIK=90+\angle IBA$, but I don't know why it is enough. Can you please mention how it finishes from there? Can someone please help me with this?

This is a great exercise for phantom points. Let $K'$ be the point on segment $FD$ such that $\angle BAF = \angle FAK'$, and let $\overline{IK'}$ intersect $\overline{AD}$ at $C'$. Since $\angle EDA = \angle EBC = \angle ECB = \angle EDB$, it follows that $E$ is the incenter of $\triangle ABD$. Now note that \[ \angle K'AD = \angle BAD - \angle BAK' = 2(\angle BAE - \angle BAI) = 2\angle FAE = \angle K'BA, \]and so $\triangle BAD \sim \triangle AK'D$. Thus by the angle bisector theorem, \[ \frac{AC'}{C'D} = \frac{AK'}{K'D} = \frac{BA}{AD} = 2, \]and so $C = C'$. This implies that $K = K'$ and we are done.

Claim 1: $E$ is the incenter of $\triangle ABD.$ We have $$\angle ADE=\angle EBC=\angle ECB=\angle EDB$$from $EB=EC$ as well as $EDCB$ being cyclic. Furthermore, $E$ clearly lies on the bisector of $\angle BAD$, so $E$ is the incenter of $\triangle ABD.$ Now, we will use barycentric coordinates, but with reference triangle $\triangle ABD$. Let $a=BD,b=AD,c=AB$ (so that $c=2b.$) Consider the circle $(AEB)$. Since it passes through $A$ and $B$, we have $u=v=0.$ Plugging in $E=(a:b:c)$ reveals that $$w=ab,$$so the equation of $(AEB)$ is $$(x+y+z)(abz)=a^2yz+b^2xz+c^2xy.$$Intersecting this with $BD$ (i.e. $x=0$) and also setting $x+y+z=1$ gives $$abz=a^2yz$$$$y=\frac{b}{a}.$$Thus, $$F=(0,\frac{b}{a},1-\frac{b}{a})=(0:b:a-b).$$Next, we use cevians $AF$ (which tells that the $y$ and $z$ coords are in a $b$ to $a-b$ ratio) and $BE$ (which tells us that the $x$ to $z$ coords are in a $a$ to $c$ ratio) to get $$I=(a(a-b):bc:c(a-b)).$$Of course, $c=(-1,0,2)$, so the equation of line $IC$ we can compute is $$u=-2bc,v=(a-b)(2a+c),w=-bc.$$We can check that these work by plugging in $I$ and $C$. This means that intersecting this with $x=0$ (line $BD$), we have that $$K=(0:bc:(a-b)(2a+c)).$$ This means that $$\frac{DK}{DB}=\frac{bc}{bc+(a-b)(2a+c)}=\frac{bc}{a(2a+c-2b)},$$which means that $$DK=\frac{bc}{2a+c-2b}.$$However, since $c=2b$, this is just $$DK=\frac{bc}{2a}=\frac{b^2}{a}.$$Which means that $$DK\cdot DB=a\cdot\frac{b^2}{a}=b^2=AD^2.$$ Thus, we have $$\triangle DAK\sim\triangle DBA.$$Let $\angle BAC=2\alpha$. From the similar triangle, we have $$\angle AKD=2\alpha$$as well. However, since $$DK\cdot DB=AD^2,$$we also have $$DK\cdot DB=CD^2,$$which means $\triangle DKC\sim \triangle DCB.$ Thus, $$\angle DKC\equiv \angle DCB=90-\alpha,$$so $$\angle IKB=90-\alpha$$and hence $IK$ bisects $\angle BKA$ (since $\angle BKA=180-\angle ATD=180-2\alpha$). Since we have already established that $E$ is the incenter of $\triangle ABD$ and hence $BI$ bisects $\angle ABK$, we are done.

Let $AB$ intersect $(BEC)$ again at $D'$ and let $AE$ intersect $(BEC)$ again at $G$. We will prove the result in two steps. First, we'll show that $BI$ bisects $\angle ABK$. Then, we'll show that $\angle AIK=90^\circ+\tfrac12 \angle ABK$. Note that since $EG$ is a perpendicular bisector of chord $BC$ of $(BEC)$, it is a diameter, so since $D$ and $D'$ are symmetric about $EG$ \[\angle ABI=\angle D'BE=\angle DBE=\angle KBI\]which proves the first part. For the second part, let $X$ be the reflection of $E$ across $D$. We have $AECX$ is a parallelogram and \[90^\circ+\tfrac12\angle ABK=90^\circ+\angle EBD=90^\circ+\angle EGD=\angle AEX\]Note that $\angle IAE=\angle FAE=\angle ABE=\angle ACE=\angle CAX$ and $\angle AXC=\angle AEC=\angle AEB=\angle AEI$ so $\triangle AXC$ is similar to $\triangle AEI$. This implies that $\triangle AEX$ is similar to $\triangle AIC$ so $\angle AEX=\angle AIC=\angle AIK$ as desired.

Solving this problem is like unscrambling a Rubic's cube... Because $E$ lies on the angle bisector of $\angle BAC$, $BE=EC$, and thus $E$ lies on the angle bisector of $\angle BDA$. So $E$ is the incenter of $BAD$ Because $\angle EFD=\angle EAB=\angle EAD$, triangles $EAD$ and $EFD$ are equal and hence $AD=FD$ Now triangles $BEC$ and $FDC$ are similar, we will only use $\angle BCE=\angle DCF$ from it. Let $AE \cap BD = M$. Now use Isogonal Lemma in $\angle BCA$ to get that $\angle ICD=\angle MCB=\angle DBC$, and so $K$ is actually the Humpty point of our triangle! But we will only use that $\angle BKA = 180-\angle BAD$. Finally, to prove that $I$ is incenter of $BAK$, we will just check that $\angle AIB=90 +\frac{\angle BKA}{2}$. Notice that triangles $AIB$ and $EFB$ are similar, and so $\angle AIB=\angle EFB=180-\frac{\angle BAC}{2}$, while $90+\frac{\angle BKA}{2}=90+90-\frac{\angle BAC}{2}=\angle AIB$, just as desired

Just take point $X$ on line $BD$, such that $AD^2= DX*DA=DC^2$. Then by angle chasing $X=K$ and the rest is also angle chasing.