Any ideas?

[asy][asy] size(10cm); draw(unitcircle); pair A = dir(70); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair K = dir(90); pair ab = dir(140); pair I_C = 2*ab-I; pair ac = dir(20); pair I_B = 2*ac-I; pair F = (5*B+6*A)/11; path cb = circle(B,6/11*2*sin(70/180*pi)); path BC = (10*B-9*C)--(10*C-9*B); pair Z = IP(cb,BC); path cc = circle(C,2*sin(120/180*pi)-6/11*2*sin(70/180*pi)); path AC = (10*A-9*C)--(10*C-9*A); pair E = IP(AC,cc); draw(I_B--F); draw(A--B--C--A--cycle); draw(I_C--K--A--I_B); draw(I_C--E); pair T = extension(I_C,E,I_B,F); path KT = (100*T-99*K)--(100*K-99*T); path KZ = (100*Z-99*K)--(100*K-99*Z); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$E$",E,dir(E)); dot("$K$",K,dir(K)); dot("$I_B$",I_B,dir(I_B)); dot("$I_C$",I_C,dir(I_C)); dot("$F$",F,dir(F)); dot("$Z$",Z,dir(Z)); dot("$T$",T,dir(T)); path unit = unitcircle; pair P = IP(KT,unit); draw(K--T--P); dot("$P$",P,dir(P)); pair Q = IP(KZ,unit); draw(K--Z--Q); dot("$Q$",Q,dir(Q)); draw(circumcircle(F,T,E)); [/asy][/asy] We first let $Z$ be the point on $BC$ such that $BZ=BF$ and $CZ = CE$. Now let $KZ$ intersect the circumcircle of $\bigtriangleup ABC$ at $Q$. Claim 1: $\bigtriangleup KBZ \sim \bigtriangleup KQB$. Proof: $\angle BKQ$ is common and $\angle KQB = \angle KCB = \angle KBC = \angle KBZ$. Now we have $$\frac{KQ}{QB}=\frac{KB}{BZ}=\frac{KB}{BF}=\frac{KI_B}{BF}$$ This rearranges to $$\frac{KQ}{KI_B}=\frac{BQ}{BF}$$ Now since $\angle FBQ = \angle ABQ = \angle AKQ = \angle I_BKQ$, thus we have $\bigtriangleup BFQ \sim \bigtriangleup KI_BQ$. Thus the spiral similarity also implies that $\bigtriangleup KQB \sim \bigtriangleup I_BQF$. $\angle TFQ = \angle I_BFQ = \angle KBQ = 180^{\circ}-\angle QPK = 180^{\circ}-\angle QPT$. Thus $TPQF$ cyclic. Similarly $TEPQ$ cyclic. The conclusion follows.

Let $I_A$ be the $A$-excenter of $\triangle ABC$, $I_BI_C$ meets $BC$ at $K'$. Consider a transformation combined by a inversion with center $A$ and radius $\sqrt{AB\cdot AC}$ and then a reflection with mirror $AI_A$. In this transformation, $B\to C, C\to B, K\to K', I_B\to I_C, I_C\to I_B$. Suppose that it sends $E\to E', F\to F'$. Let the circumcircles of $\triangle I_CF'A$ and $\triangle I_BE'A$ meets at another point $T'$, the circumcircle of $\triangle AK'T'$ meets $BC$ at another point $P'$, lines $I_BE'$ meets $I_CF'$ at $Q$. By Pappus' Theorem, $Q$ is on $BC$. Because $\angle QP'T=\angle K'AT'=\angle I_BE'T=\angle QE'T$, $Q,P',E',T'$ are concyclic. For the same reason, $Q,P',T',F'$ are concyclic, therefore $Q,P',E',T',F'$are concylic. $\square$

One of the hardest geometry problems in Chinese contests that are below CMO. I just realized from #4 that the sine bash in the claim can be replaced by a one-line Pappus proof.

Hard problem. This was too good. We have that $BF+ CE=BC$ so $AEIF$ is cyclic (well-known). Applying $\sqrt{bc}$-inversion with reflexion in the bisector of $\angle BAC$ we have the follow problem: Quote: In a triangle $\triangle ABC$, let $I_A,I_B$ and $I_C$ be the excenters of $\triangle ABC$ opposite $A,B,C$ respectively. Let $E' \in AC$ and $F' \in AC$ such that $I_A \in E'F'$. Let $K'=I_CI_B \cap BC$, $T' = \odot (I_BE'A) \cap \odot (I_CF'A) \neq A$ and $P'= \odot (AT'K') \cap BC \neq K'$. Show that $T',F',P',T'$ are concyclic. Let $X= I_CF' \cap I_BE'$ by pappus theorem (In $I_C-A-I_B$ and $E'-I_A-F'$) we have $X \in BC$. Later we have that: $$\angle T'P'X= \angle T'KA = \angle TKI_B = 180 - \angle T'E'X$$So we have $T'P'XE'$ is cyclic, analogously $T'P'XF'$ is cylic so $T'P'E'F'$ is cyclic, as desired. $\blacksquare$

Let $I_BAF$ meet $ABC$ at $S$. Note that $\angle AI_BS = \angle BFS$ and $\angle I_BSF = \angle KAB = \angle KSB \implies \angle FSB = \angle I_BSK$ so $FSB$ and $I_BSK$ are similar so $\angle TPS = \angle 180 - \angle KBS = \angle 180 - \angle I_BFS = \angle 180 - \angle TFS$ so $TFSP$ is cyclic. we will do the same approach for $I_CAES$ and will prove $TESP$ is cyclic with same approach so $TEFSP$ is cyclic. we're Done.