I have some useful lemmas ,but i have't finished. Any full solution?

The only solution I came up with is complex numbers bashing. Let $X,Y,Z,T$ be midpoints of arcs $AB,BC,CD,DA$ that do not contain other points. $X',Y',Z',T'$ are midpoints of the other arcs $AB,BC,CD,DA$. Now $k$ is a unit circle, $A = 1, X = a, B = a^2, Y = a^2 b, C = a^2 b^2, Z = a^2 b^2 c, D = a^2 b^2 c^2, T = -abc$. We find the incircles by formula for chord intersection. One can show that these 2 lines through incircles pass through $X$ and $Z'$ respectively. After that we use formula $q = \frac{\omega - p}{1 - \bar{\omega}p}$ where $p,q$ are ends of a chord and $\omega$ is a point on this chord. First, we set $(p, \omega)$ as $(X, I_{ABE})$ and later as $(Z', I_{CDE})$ and we get two equal $q$, so it is what we wanted. All in all, it looks awfully and it is awful indeed.

Thank you Kaszubki for nice solution.Any another solution(without complex plane and coordinata geometry ).I'm waiting...

Hi Sardor do you mind sharing your lemmas? This problem seems to be very difficult.

Well it all suffices to prove that the first incenter line goes through the midpoint of arc $AB$ and the second incenter line goes through the midpoint of arc$DAC$. My approach had been working with isogonals and I ended up deducing the property above. I will post full solution soon Solution: WLOG $A,B$ are on segments $DF,CF$ and $\angle D>\angle C$. Let $I_1,I_2,I_3,I_4$ denote the incenters of $ABE,CDE,ABF,CDF$. Let $M,N$ denote the midpoints of arcs $AB,DAC$. We will first prove that $I_2,I_4,N$ are collinear, symmetric result $I_1,I_3,M$ can be proven analogously. Some angle chasing reveals that $\angle DCI_2=\angle NDI_4, \angle I_2DC=\angle NCI_4$. Construct the isogonal conjugate of $I_2$ wrt $NDC$ denoted by $I_2'$. Hence $I_4,I_2'$ are two symmetric points about the perpendicular bisector of $DC$, which $N$ lies on. Hence $\angle CNI_2=\angle CNI_2'=\angle DNI_4$ $\Rightarrow I_2,I_4,N$ are collinear. With these two results in mind, it now suffices to show that $\angle (I_2I_4,I_1I_3)=\frac {\angle D-\angle C}{2}$. Let $I_{12}$ denote the reflection of $I_2$ over the midpoint of $DC$. It's a well known lemma that if $\angle I_4CI_2=\angle I_4DI_2$ which is true by angle chasing, then $I_4I_2,I_4I_{12}$ are isogonals wrt $\angle DI_4C$. Since $I_4DI_{12}C\sim I_3BI_1A$, therefore $\angle DI_4I_2=\angle CI_4I_{12}=\angle AI_3I_1$ $\Rightarrow \angle (I_2I_4,I_1I_3)=\angle (DI_2,A_3)=\frac {\angle D-\angle C}{2}$ and we are done.

My sol is similar to XML's. We note that $N, I_2, I_4$ are collinear, then we can show that the pencils through $N, M$ are projective, so if we fix $M, N$ by moving $B, C$, the intersections lie on a conic passing through $M, N$. We define this conic as $ABCDMN$, i.e. the circumcircle.

Turkey 2003 2nd $Q 5$ will be the useful Lemma

I think I have gotten to the bottom of the configuration. Theorem 1 (Sawayama's lemma). Consider a cyclic quadrilateral $ABCD$. Let $I=I_{ABC}$ be its incenter, and suppose the circle $m$ tangent to $AC$, $BD$ and the arc $AB$ not containing $C,D$ touches $AC$ at $X$, $BD$ at $Y$, and circle $k=\odot(ABCD)$ at $T$. Then $I$ lies on $XY$, and quadrilaterals $TBYI$, $TAXI$ are cyclic. Proof. First, let $M=M_{AC}$ be the midpoint of arc $AC$, and $I'=BM\cap XY$. Consider the homothety from $T$ between $m$ and $k$: due to parallel tangents, $X$ maps to $M$, and so arc $TX$ maps to arc $TM$, and thus they are of equal length, whence \[ \sphericalangle TYI'=\sphericalangle TYX = \sphericalangle TBM= \sphericalangle TBI', \]and so $TBYI'$ is cyclic. Second, note that $MX\cdot MT=MA^2=MC^2$ because an inversion at $M$ with radius $MA=MC$ maps $AC$ to $k$, so it switches $X$ and $T$. Additionally, $MA=MC=MI$, as is well-known. Thus $MI'^2=MX\cdot MT$ would imply $I=I'$ and we'd be done. But $\triangle MXI'\sim\triangle MI'T$, because \[ \sphericalangle MI'T=\sphericalangle BI'T=\sphericalangle BYT=\sphericalangle DYT= "\sphericalangle YYT"=\sphericalangle YXT=\sphericalangle I'XM. \quad \blacksquare \] Theorem 2. Using the same notation, if $E=AC\cap BD$, $F=BC\cap AD$ and $H,I,J,K$ are the incenters of triangles $ABD,ABC,ABE,ABF$, then $T,J,K$ are collinear. Proof. Note that $\sphericalangle BJA=\sphericalangle BYI$ - both can be realized as the same directed angle of size $\pm\left(90^\circ-\frac 12 \theta\right)$ where $\theta=\angle AEB$. So $\sphericalangle BJI=\sphericalangle BYI$ and by Theorem 1, $B,T,Y,I,J$ all lie on a circle $\omega_B$. Similarly, $A,T,X,H,J$ lie on a circle $\omega_C$. Consider the radical center of $\omega_B$, $\omega_C$ and $\odot(ABIH)$ (the latter is centered at the midpoint of smaller arc $AB$). It is $K$, because $K=BI\cap AH$: $BI$ bisects $\angle ABE$ and $AH$ bisects $\angle BAE$. Therefore, $K$ lies on the radical axis of $\omega_B$ and $\omega_C$, which is $TJ$. $\blacksquare$ The exact same proof tells us that the incenters of $\triangle CDE,\triangle CDF$ also are collinear with $T$, and so we are done with the proposed problem.

Bulgaria 2014/6 wrote: Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$. Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$. Proposed by N. Beluhov Excellent problem! Let $P$ be the touching point of the curvilinear incircle of triangle $ECD$ with the circumcircle of quadrilateral $ABCD$. I claim that $P$ lies on both these lines. We'll show only one case here, namely for the line joining incenters of $\triangle ABE$ and $\triangle ABF$, which we shall label as $I, J$ respectively. The other case can be handled in the exact same fashion. Let $M$ be the midpoint of arc $AB$ not containing either of $C, D$. Note that this implies $P, I, M$ are collinear. To end our proof, consider the following Claim: $M, I, J$ are collinear. (Proof) Complete isosceles trapezoid $ABKI$ with $AB \parallel KI$. Let $\angle ECD=\alpha, \angle BCD=\beta$ and $\angle MAB=\angle MBA=\gamma$. Notice that $$\angle KAB=\angle IBA=\frac{1}{2} \alpha=\frac{1}{2}\left(\beta-\gamma\right)=\angle MAJ,$$and similarly $\angle MBJ=\angle KBA$ so $K, J$ are isogonal conjugates in $\triangle MAB$. Evidently, $MI, MK$ are isogonal in $\angle AMB$ so our claim is valid. $\square$

Oops I had too much fun with Pascal's. Let the incenters of $ABE, ABF, CDE, CDF$ be $I_1, J_1, I_2, J_2$. Let $M$ and $N$ be the midpoints of arcs $AB$ and $CAD$. Consider the circle with center $M$ passing through $A$ and $B$. Then by Pascal on the the following 6 points: $A$, incenter of $ABC$, $C$-excenter of $ABC$, $B$, incenter of $ABD$, $D$-excenter of $ABD$ We get that $I_1, J_1, M$ are concurrent. Similarly, $I_2, J_2, N$ are concurrent. Let $L$ be the midpoint of arc $BC$ and the intersection point be $T$. By the converse of Pascal's on $ALDMTN$, and because $AN$ intersects $DM$ at the $D$-excenter of $ADE$ (and therefore lies on $EI_1I_2$), those six points lie on some conic. But five points already lie on the circle $k$ so $T$ must also lie on this circle.

Here is a solution found during Thailand TST group's live solve. We recall the following known results. Lemma 1: [Sawayama] Let $\triangle ABC$ be a triangle and $D$ be a point on segment $BC$. Suppose that a circle $\omega$ tangents to $AD$ at $E$, $CD$ at $F$ and $\odot(ABC)$ internally at $T$. Then $EF$ pass through incenter of $\triangle ABC$. Lemma 2: [Protassov] Let $\triangle ABC$ be a triangle with incenter $I$ and $\omega$ be a circle through $B$ and $C$. Suppose that the circle tangents to $AB$, $AC$ and $\omega$ internally at $T$, then $TI$ bisects $\angle BTC$. Now, construct the circle $k_1$ tangents to $AD, BC$ and $k$ internally at $T_1$. Construct the circle $k_2$ tangents to $AC, BD$ and $k$ internally at $T_2$. Suppose that both $T_1,T_2$ are on the minor arc $CD$, we use Lemma 1 to prove the following (which I think is also known). Claim: $T_1=T_2=T$ Proof: Let $I_A, I_B$ be the incenters of $\triangle ACD$ and $\triangle BCD$. Let $\omega_1$ tangent to $AD, BC$ at $X_1, Y_1$ and let $\omega_2$ tangent to $AC, BD$ at $X_2,Y_2$. By Lemma 1, we deduce that $X_1$, $Y_1$, $X_2$, $Y_2$, $I_A$, $I_B$ are colinear. By homothety, we see that $T_1X_1$ and $CI_A$ meet on a point $K\in k$. Moreover, lines $T_2X_2$ and $DI_A$ meet on a point $L\in k$. Thus by Pascal's theorem on $T_1LDACK$, we get that $T_1, L, X_2$ are colinear hence $T_1=T_2$. $\blacksquare$ Now the problem falls apart by Lemma 2. The lines are angle bisector of $\angle CTD$ and $\angle ATB$ and they meet at $T$.

MarkBcc168 wrote: Here is a solution found during Thailand TST group's live solve. We recall the following known results. Lemma 1: [Sawayama] Let $\triangle ABC$ be a triangle and $D$ be a point on segment $BC$. Suppose that a circle $\omega$ tangents to $AD$ at $E$, $CD$ at $F$ and $\odot(ABC)$ internally at $T$. Then $EF$ pass through incenter of $\triangle ABC$. Lemma 2: [Protassov] Let $\triangle ABC$ be a triangle with incenter $I$ and $\omega$ be a circle through $B$ and $C$. Suppose that the circle tangents to $AB$, $AC$ and $\omega$ internally at $T$, then $TI$ bisects $\angle BTC$. Now, construct the circle $k_1$ tangents to $AD, BC$ and $k$ internally at $T_1$. Construct the circle $k_2$ tangents to $AC, BD$ and $k$ internally at $T_2$. Suppose that both $T_1,T_2$ are on the minor arc $CD$, we use Lemma 1 to prove the following (which I think is also known). Claim: $T_1=T_2=T$ Proof: Let $I_A, I_B$ be the incenters of $\triangle ACD$ and $\triangle BCD$. Let $\omega_1$ tangent to $AD, BC$ at $X_1, Y_1$ and let $\omega_2$ tangent to $AC, BD$ at $X_2,Y_2$. By Lemma 1, we deduce that $X_1$, $Y_1$, $X_2$, $Y_2$, $I_A$, $I_B$ are colinear. By homothety, we see that $T_1X_1$ and $CI_A$ meet on a point $K\in k$. Moreover, lines $T_2X_2$ and $DI_A$ meet on a point $L\in k$. Thus by Pascal's theorem on $T_1LDACK$, we get that $T_1, L, X_2$ are colinear hence $T_1=T_2$. $\blacksquare$ Now the problem falls apart by Lemma 2. The lines are angle bisector of $\angle CTD$ and $\angle ATB$ and they meet at $T$. Can't we just finish after Lemma 2? Defining T as the tangency point of a circle tangent to segments $ED$, $EC$ and internally to $k$, we can use the lemma to deduce that the incenters of $\triangle AEB, \triangle AFB$ lie on the bisector of $\angle ATB$. Similarly, the other two incenters lie on the bisector of $\angle DTC$. So, the lines and the circle $k$ concur at $T$. Am I wrong? edit: yes, I am wrong