Let $AL$ and $BK$ be the angle bisectors in a non-isosceles triangle $ABC,$ where $L$ lies on $BC$ and $K$ lies on $AC.$ The perpendicular bisector of $BK$ intersects the line $AL$ at $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK.$ Prove that $LN=NA.$
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Tags: geometry, circumcircle, perpendicular bisector, angle bisector, geometry proposed
hyperbolictangent
02.12.2012 23:03
Since the perpendicular bisector of $BK$ and the angle bisector of $\angle{KAB}$ concur on the circumcircle of $\triangle{KAB}$, quadrilateral $ABMK$ is cyclic. Then quadrilateral $ABLN$ is cyclic as well, since $\angle{NLA} = \angle{AMK} = \angle{ABK} = \angle{ABL}$. But since $N$ is on the angle bisector of $\angle{ABL}$ and simultaneously on the circumcircle of $\triangle{ABL}$, it must be the midpoint of $\widehat{AL}$, whence it lies on the perpendicular bisector of $AL$ and so $NL = LA$.
Potla
02.12.2012 23:16
Let $D,E$ be feet of perpendiculars from $M$ onto $AB$ and $AC,$ respectively. Let $I=AL\cap BK.$
Since $AM$ bisects $\angle DAE,$ so $MD=ME.$ Also, $M$ lies on the perpendicular bisector of $BK$ so $MB=MK.$ Then, note that $MBD\cong MEK,$ leading to $\angle MKE=\angle MBD.$ This leads to $\angle MBA+\angle MKA=\pi,$ so that $MBAK$ is cyclic. Therefore $\frac{IA}{IB}=\frac{IK}{IM};$ and since $ILN\sim IMK,$ so we have $\frac{IK}{IM}=\frac{IN}{IL}.$ Thus $IA\cdot IL=IB\cdot IN,$ thus $L,A,B,N$ are concyclic. So, $\angle LAN=\angle LBN=\angle ABN=\angle ALN$ since $BN$ bisects $\angle ABL.$ So, $LAN$ is isosceles, and $LN=NA.\Box$
cmtappu96
03.12.2012 07:06
hyperbolictangent wrote: Since the perpendicular bisector of $BK$ and the angle bisector of $\angle{KAB}$ concur on the circumcircle of $\triangle{KAB}$, . This was problem no 1 of RMO-2, 2011. It is pretty nice that they give sequels to their problems
thecmd999
21.06.2013 02:48
Stolen from the 2010 JBMO :O
jayme
08.11.2014 14:00
Dear Mathlinkers, for proving that ABLN is cyclic, we can use a convese of the Reim's theorem. Sincerely Jean-Louis