Let $AD$ be an internal angle bisector in triangle $\Delta ABC$. An arbitrary point $M$ is chosen on the closed segment $AD$. A parallel to $BC$ through $M$ cuts $AB$ at $N$. Let $AD, CM$ cut circumcircle of $\Delta ABC$ at $K, L$, respectively. Prove that $K,N,L$ are collinear.
Problem
Source: Serbian JBMO TST 2018, problem 1
Tags: geometry
enhanced
22.05.2018 01:04
Too easy ! By Pascal theorem on $BAKKLC$ $\implies$ the line joining $ AB\cap KL,M$ is parallel to $BC$ $\implies AB\cap KL=N$ $\blacksquare$
DanDumitrescu
22.05.2018 01:05
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.
jayme
22.05.2018 14:00
Dear Mathlinkers, the last proof involves angles chasing for the cocyclicity Then we finish with the Reim's theorem... Sincerely Jean-Louis
Mahdi_Mashayekhi
01.01.2022 13:48
∠ANM = ∠ABC = ∠ALM ---> AMNL is cyclic. ∠KLC = ∠KAC = ∠NAM = ∠NLC ---> K,N,L are collinear.
tooral888
03.06.2023 13:38
........