# Geometry problem involving triangle and three intersecting circles

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.