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? enter image description here


1 Answer 1


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.


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