Let $M$ be the midpoint of the side $BC$ of triangle $ABC$. The bisector of the exterior angle of point $A$ intersects the side $BC$ in $D$. Let the circumcircle of triangle $ADM$ intersect the lines $AB$ and $AC$ in $E$ and $F$ respectively. If the midpoint of $EF$ is $N$, prove that $MN\parallel AD$.
Problem
Source: Mongolian MO 2007 Grade 11 P1
Tags: geometry
hakN
09.04.2021 12:22
Can anyone post a non-bashy solution for this?
top10anime
09.04.2021 14:31
Let $X$ be the midpoint of arc $BC$ without $A$, and $U$, $V$ feet from $X$ onto $AB$ and $AC$. Obviously $UV \perp AX$. By chasing $AD$ is the internal bisector of $\angle EAF$ and thus $AX$ is the external bisector of $\angle EAF$ so $XN \perp EF$. Now $N-U-V$ and $M-U-V$ are Simson lines from $X$ wrt. $\triangle AEF$ and $\triangle ABC$ so $MN \perp AX$.
hakN
09.04.2021 14:38
Wonderful solution! Thanks @top10anime
zuss77
09.04.2021 14:40
I see I'm late. Let $PQ$ - diameter of ($ABC$) with $Q\in AD$. Then $DP$ - diameter of $(ADM)$ and $P$ - midpoint of the arc $EF$. So $P$ - center of spiral similarity $QBCM\mapsto DFEN$. Hence $PM/PQ=PN/PD$.
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