Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.
Problem
Source: 2011 Balkan Shortlist G2 BMO
Tags: geometry, symmedian, equal angles, circles, circumcircle, angle bisector
amar_04
03.04.2020 19:32
parmenides51 wrote: Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal. Claim:- $M\in\odot(ALO)$ So this suffices to prove that $\angle OAL=90^\circ$ which further suffices to prove that $\angle EAL=\angle ALC=\angle OAD$. But as $A-$ Appolonius Circle $\perp\odot(ABC)$. Hence, it must hold true. Hence, $M\in\odot(ALO)$. Now as $AO=NO$. Hence, $ML$ is the internal bisector of $\angle AMN\implies LA=LN$. But $LA$ is tangent to $\odot(ABC)$. So, $ABNC$ is a harmonic quadrilateral. Hence, $AN$ is the $A-$ Symmedian of $\triangle ABC$ $\blacksquare$
Kimchiks926
29.10.2021 19:52
Observe that $(E,D;B,C)=-1$. Since $L$ is midpoint we have $LD=LE=LA$ moreover:
$$ LE^2 = LB \cdot LC = LA^2$$This means that $LA$ is tangent to $\odot(ABC)$. Consequently, $\angle LAO = \angle LMO=90^{\circ}$, implying that $LAOM$ is cyclic quadrilateral. This is enough to conclude since we have $\angle ONL=90^{\circ}$, meaning that $LN$ is tangent to $\odot(ABC)$ too. Therefore $ABNC$ is harmonic quadrilateral, which gives us that $AN$ is symmedian. Thus $\angle BAN = \angle CAM$ as desired.