Let $\triangle ABC$ be a triangle, and $\Gamma$ be a circle with center $O$ passing through $A$, which intersects $[AB]$ in $K$, $[BC]$ in $L$ and $M$ such that $L$ is between $B$ and $M$, $[AC]$ in $N$. Let $U$ be the center of the circle circumscribed of $\triangle KBL$ and $V$ be the center of the circle circumscribed of $\triangle NCM$.
How can we show that $(UL)$ and $(VM)$ intersect on $\Gamma$ ?
I tried, but I can't quite express the fact that the intersection of $(UL)$ and $(VM)$ is on $\Gamma$, except that the power of this point relative to $\Gamma$ is zero, or that the intersection of $(UL)$ and $\Gamma$, $V$ and $M$ are aligned. However, I can't find any properties involving this. I have also tried to deal with the problem analytically, but I don't think it is particularly relevant, because there is no obvious orthonormal marker that could be used to write the equations of circles, for example.
Could you help me?
Suppose $UL$ and $VM$ intersects at $D$.
The strategy is to show that $\angle OAL=\angle DLM$ and $\angle OAM=\angle DML$.
If it is done, then we have \begin{align} \angle MDL & =180^{\circ}-\angle DLM-\angle DML &\\ & =180^{\circ}-(\angle OAL+\angle OAM) &\\ & =180^{\circ}-\angle LAM & \end{align} This implies that $A$, $L$, $D$ and $M$ are concyclic.
To show that $\angle OAL=\angle DLM$:
Let $\angle OAL=x$.
Since $OA=OL$, $\angle OLA=\angle OAL=x$.
Since $OKUL$ is a kite with $OK=OL$ and $UK=UL$, $OU$ bisects $\angle KOL$ and $\angle KUL$.
Then we have $\angle UOL=\frac{1}{2}\angle KOL=\angle KAL$.
Similarly, we have $\angle OUL=\frac{1}{2}\angle KUL=\angle KBL$.
Since $\angle OLU=180^{\circ}-\angle UOL-\angle OUL$ and $\angle ALB=180^{\circ}-\angle KAL-\angle KBL$, $\angle OLU=\angle ALB$. Hence, we have $\angle ULB=\angle OLA=x$.
Then note that $\angle DLM=\angle ULB=x$. Finally, we have $\angle OAL=\angle DLM$.
The result that $\angle OAM=\angle DML$ can be obtained similarly.