Totally similar to this ELMO problem: https://artofproblemsolving.com/community/c6h1262190p6556898 Anyone who have seen that one can finish this P7 in ten minutes

Let $J$ be the intersection of $LB,KC$, $Y',Z'$ is the intersection of $BK,CL$ with $(OBC)$. Then, we have: $\angle{BLZ'}=\angle{Z'CB}=\angle{Z'JB}$, thus $Z'$ lies on the perpendicular bisector of $LJ$. Similarly, we have $Y'$ lies on the perpendicular bisector of $JK$. Thus, if we can obtain that $X$ is the center of $(JLK)$ then the problem will be solved. $\angle{BJC}=180^{\circ}-\angle{JBC}-\angle{JCB}=180^{\circ}-(180^{\circ}-2.\angle{ABC})-(180^{\circ}-\angle{ACB})=180^{\circ}-2.\angle{BAC}$. Thus $J$ lies on $(BOC)$. Let $I$ be the intersection of tangent of $B,C$ of $(ABC)$. From a bit of angle changing and $IB=IC$,$LB=BC=KC$ we can prove that $\triangle LBI=\triangle KCI$, thus $\angle{LIK}=\angle{LJK}$ and $(LIJK)$ is concyclic.. Let $X'$ be the center of $(LIJK)$, $O_1,O_2$ be the center of $(ABL),(ACK)$. It's easy to prove that $\triangle ABL=\triangle AKC$. From angle changing, we can prove that $\angle{O_1LX'}=\angle{O_2KX'}$ (as they are equal to $\angle {ILB} $and $\angle{IKC}$). Thus X' lies on the radical axis of $(ALB)$ and $(AKC)$ (which is line AH with H is the orthocenter). Thus $X '\equiv X$

I have find something in the graph In triangle ABC, the outer center is O, make the outer circle of triangle BOD, and the circle T intersects AB and AC with E and F respectively. BD intersects CE at F and FT, and the height on the side of BC (in the triangle ABC) intersects G Proof: AEFG four points are circular

Extend $BL$ and $CK$ to meet at $M$. Let $D,E,F$ be the foot of perpendicular from $A$ onto $BC,BL,CK$ respectively. Then notice that $A$ is the $M$-excentre of $\triangle MBC$. Hence we construct $E_C$ and $E_B$, the excentres of $\triangle MBC$ opposite $B$ and $C$ respectively. Let the ex-touch points on segments $BM$ and $MC$ be $P$ and $Q$ respectively. Then $AD, E_CP,E_BQ$ are concurrent at point $T$ (which is also the circumcenter of triangle $AE_BE_C$) Notice that since $BL=BC=CK$, $BE=BD$,$CD=CF$, we have $KF=BD=QM$ and $EL=CD=PM$. Hence the perpendicular bisector of $ML$ is the same as that of $PE$, and the perpendicular bisectors of $MK$ is the same as that of $QF$. Since both of them will intersect on the midpoint of $AT$, we can conclude that the circumcenter of $\triangle KLM$ is the midpoint of $AT$. Since $X$ is on $AT$ such that $XL=XK$, $X$ is this circumcenter, and $XY$ and $XZ$ are the perpendicular bisectors of $MK$ and $ML$ respectively. As such, $\angle ZMB=\angle ZLB=\angle CLB=\angle BCZ$ and $\angle YMC=\angle YKC=\angle YBC$. This gives $B,C,Y,Z,M$ concyclic. Notice that $O$ lies on the circumcircle of $BMC$ and we are done.

Hello! Thanks for your decision. It`s wonderful! I solved this problem in exactly the same way, but I could not prove that circumcenter of triangle KLM lies on AH (AT). But you helped) Thank you & Good luck!