a very cute problem using the result obtained here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827187sid=864d49327d5df0a637f8725f4205e6a2#p827187 BMO 2007 we get that OE=OF where O is the intersection of the diagonals of quadrilateral EBCF CBF=g(angle)=CFB and BEC=BCE=b from triangle ABC we get b+g=60 but OEF=OFE=f so f=30 from trianlge AEF so KAC=f let us now prove OCFK is cyclic we have from angle chasing COF=60(OBAC-cyclic) and CKF=60(EACK cyclic) so OCFK cyclic so ECK=EAK=KAC so K is on the bisector now let us take the case when EC=BF which is omited in the BMO 2007 problem then BCF and BEC are congruent and b=g so BO=OC and one more time OE=OF so we continue with the same thing from above

Denote $AB=x$, $AC=y$, $BE=CF=BC=z$. By cosine law applied in triangle $ABC$, $z^2=x^2-xy+y^2$. (1) By cosine law in triangle $AFE$, \[EF^2=(x+z)^2-(x+z)(y+z)+(y+z)^2\]. Using (1), \[EF=\sqrt{z(x+y+2z)}\]. As $A,C,K,E$ lie on one circle, $FK\cdot {FE}=FC \cdot{FA}$, from which $FK=\frac{(y+z)\sqrt{z}}{\sqrt{x+y+2z}}$. So, $EK=EF-FK=\sqrt{z(x+y+2z)}-\sqrt{\frac{z}{x+y+2z}}\cdot{(y+z)}=\frac{(x+z)\sqrt{z}}{\sqrt{x+y+2z}}$. It means that $EK\cdot{EF}=z(x+z)$. Finally, $\frac{EK}{KF}=\frac{EK\cdot{EF}}{KF\cdot{EF}}=\frac{z(x+z)}{z(y+z)}=\frac{x+z}{y+z}=\frac{AE}{AF}$, which means that $K$ lies on the bisector of $\angle{A}$, as desired.

my solution : first I'll prove a lemma : in a quadrilateral $ABCD$, if we have $\angle DAC=\angle BAC, BC=CD$ and $AB\neq AD$, then $ABCD$ is concyclic. $proof$: $\triangle ABC:\frac{AC}{sin\angle B}=\frac{BC}{sin\angle CAB}$ $\triangle ACD:\frac{AC}{sin\angle D}=\frac{CD}{sin \angle DAC}$ $BC=CD,\angle DAC=\angle BAC$ so we have $sin\angle B=sin\angle D$. because $AB\neq AD$ we have $B+D=180$ and so $ABCD$ is concyclic. now my solution to the question: suppose that $K$ is not on the internal bisector of $A$, so internal bisector of $A$ meets circumcircle of $ACE$ at $T\neq K$. $AT$ is bisector, so we have $ET=TC$. $ET=TC$ $TB=TB$ so $\Longrightarrow$ $\triangle BCT=\triangle BET$ $\Longrightarrow$ $\angle BCT=\angle BET$ $BE=BC$ $ACTE$ is concyclic, so we have $\angle BET=\angle TCF=\angle BCT$. $BC=CF$ $\angle BCT=\angle TCF$ $\Longrightarrow$ $\triangle BCT=\triangle FCT$ $\Longrightarrow$ $BT=TF$. $CT=CT$ $BT=TF$, $\angle BAT=\angle TAF$, $AB\neq AF$ so by lemma we have $ABTF$ is concyclic. $\triangle BCT=\triangle FCT$ so $\angle CFT=\angle CBT$ $ABTF$ is concyclic, so we have $\angle CBT=\angle EBT=\angle CFT$ $\angle A=60$ so $\angle B+\angle C=120$ so $\angle EBC+\angle FCB=240$ $\angle EBT+\angle CBT+\angle BCT+\angle FCT=240$ $\angle CBT=\angle EBT$ $\angle BCT=\angle FCT$ so we have $\angle BCT+\angle CBT=120$ and hence $\angle BTC=60$ $\triangle EBT=\triangle CBT=\triangle CFT$ so we have $\angle BTE=\angle BTC=\angle FTC=60$ so $\angle ETF=180$ this is a contradiction because we supposed that $T$ is not $K$ , and so $T$ is not on $EF$. Dear Mathlinkers, please read it carefully and tell me if I've made a mistake or not. I really need your answer, my life depends on this! best regards

It is very easy! we just need to prove that $K=I_a$ where $I_a$ is center of A_excircle.but $BI_aC=60$ and form that $I_a$ lies on $EF$ and $ACI_aE$ is cyclic and we are done. This is the method that i used in contest and solve it in less than 15 minutes.

I know that, but please tell me where of my solution is wrong, because when they corrected my paper, they gave me 6 out of 7!!!

please! tell me what's wrong with my solution!

It is well known that, for a triangle $\triangle AEF$ with $\angle A=60^\circ$, $FO$ and $EO$ intersect $AE$ and $EF$ respectively at $B$ and $C$, so that $BE=BC=CF$, where $O$ is the circumcenter of $\triangle ACE$, hence $\angle OEF=\angle OFE=30^\circ$, i.e. $\angle CEF=30^\circ$, or $EF$ intersects the circle $\odot (ACE)$ at $K$, the midpoint of the arc $CE$, done. A second method may use direct calculation, finding $EF$ from cosine theorem, then from $AF\cdot CF=EF\cdot KF$ finding $KF$ and, finally, $EK$ and proving $\frac{EK}{FK}=\frac{AE}{AF}=\frac{a+c}{a+b}$. Yet a 3rd method: calculating $BC$ and $EF$ by cosine formula we get $EF^2=a\cdot (2a+b+c)$, where $a, b, c$ are the sizes of $BC, CA, AB$. Taking $M$ on $FA$ produced so that $AM=AE$ we get $EF^2=CF\cdot FM$, and $EF$ is tangent to $\odot (ECM)$, with $\angle FEC=30^\circ$. Best regards, sunken rock

Using vector rotation. Denote $O$ be the center of the rotation: $\vec{EB} \mapsto \vec{CF}$ We have $A,C, O, E$ are concyclic and $A, B, O, F$ are concyclic and $\angle BOF = \angle EOC = 120^0$ From $BC = BE = CF$, hence $OB, OC$ are perpendicular bisectors of $CE$ and $BF$, respectively. It following $E, O, F$ are colinear, so $O \equiv K$. And it easy to prove $AO$ be the bisector of $\angle A$

Take $M$ midpoint of arc $EAF$ of the circle $(EAF)$; then $\Delta EFM$ is equilateral, $AEFM$ cyclic, $\angle AEM=\angle AFM, \Delta BEM=\Delta CFM$ (s.a.s), hence a $60^\circ$ rotation around $M$ maps $F$ to $E$ and $C$ to $B$, or $\Delta BCM$ equilateral, i.e. $BM=MC=BC=BE=CF$; consequently $EFMC$ is a kite, i.e. $CE$ is angle bisector of $\angle FEM$, or $\angle CEK=30^\circ$, hence $AK$ is angle bisector. Best regards, sunken rock

my proof = since $ACKE$ are cyclic. thus $\angle EKC=120$ also due to $BE=EC=CF$ we get $\angle BEC=\angle BCE=B/2$ and $\angle CBF=\angle CFB=C/2$ now using sine law in triangle $CFE,BFE$ and using $BE=CF$ with letting $\angle CEF=x$ we get $\frac{sinx}{sin(60-x)} = \frac{sin(120-C/2)}{sin(120-B/2)}$ now using $\angle B+C = 120$ and expanding the above expression using $sin(A+B)=sinAcosB+sinBcosA$ we get $2sinx=(\sqrt 3)cosx - sinx$ or $tanx=\frac{1}{\sqrt 3}$ and hence $\angle x = 30$ thus $\angle CEK=\angle KAC=30=\angle KAB$ or $AK$ bisect $\angle BAC$ we are done

Let $AB = c, AC = b, BC = BE = CF = a$. Note that $a^2 = b^2 + c^2 - bc$ and $EF^2 = (a+b)^2 + (a+c)^2 - (a+b)(a+c) = a^2 + b^2 + c^2 + ab + ac - bc = 2a^2 + ab + ac = a(2a+b+c)$ and $FK.FE = a(a+b)$ so $EK.EF = a(2a+b+c) - a(a+b) = a(a+c) \implies ABKF$ is cyclic. $\angle BEK = \angle FCK$ and $CFK = \angle EBK$ and $AB = CF$ so $BEK$ and $FCK$ are congruent so $CK = EK$ so $\angle EAK = \angle KAC$ which implies $AK$ is angle bisector of $EAF$.