Too easy ! By Pascal theorem on $BAKKLC$ $\implies$ the line joining $ AB\cap KL,M$ is parallel to $BC$ $\implies AB\cap KL=N$ $\blacksquare$

Consider $N'=KL \cap AB $ we have $LN'A=KAC=N'AM $ so LN'MA is cyclic so $ MN'A=MLA=CLA=CBA=B $so $N'M \parallel BC $ so N'=N so K,N,L are collinear.

Dear Mathlinkers, the last proof involves angles chasing for the cocyclicity Then we finish with the Reim's theorem... Sincerely Jean-Louis

∠ANM = ∠ABC = ∠ALM ---> AMNL is cyclic. ∠KLC = ∠KAC = ∠NAM = ∠NLC ---> K,N,L are collinear.

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