Let $ABC$ be a acute-angled triangle with centroid $G$, the angle bisector of $\angle ABC$ intersects $AC$ in $D$. Let $P$ and $Q$ be points in $BD$ where $\angle PBA = \angle PAB$ and $\angle QBC = \angle QCB$. Let $M$ be the midpoint of $QP$, let $N$ be a point in the line $GM$ such that $GN = 2GM$(where $G$ is the segment $MN$), prove that: $\angle ANC + \angle ABC = 180$
Problem
Source: Rioplatense Olympiad 2001
Tags: geometry, angle bisector
georgeado17
28.02.2018 15:14
My solution: Let $T$ be the midpoint of $AC$ and let $BD$ and $NT$ intersect at point $K$ then in $\Delta BNK$ note that $BG:GT=2:1$ and $GN:GM=2:1$ $\implies$ $G$ is centroid of $\Delta BNK$ too.We have:$KT=TN$ and $AT=TC$ $\implies$ quadrilateral $ANCK$ is parallelogram.Also by centroid $G$ have that $PM=MK$ $\implies$ $BP=AP=KQ=QC$.Let $\angle ABD=\angle PAB=\angle PBC=\angle QCB=\alpha$ we have that $\angle KQC=2\alpha=\angle APK$ also $PQ+QK=PQ+BP=QC$ and $QK=BP=AP$ $\implies$ $\Delta QKC=\Delta APK$.$\angle BCN=x$ ($N$ is inside interior of triangle ABC,but anyway it doesn't change solution) $\implies$ $\angle NCQ=\alpha-x$ and let $\angle QCD=y$ also let $\angle TKC=\angle TKA=m+n$ ($\angle TKD=n$ and $\angle DKA=m$) then have that $\angle PKA=\angle QCK$ which implies that $90-m-n+y=m$ $\implies$ $2m+n=90+y$ we need to prove that $\angle ANC+\angle ABC=180$ or in this case $\angle AKC+\angle ABC=180$ $\implies$ we should prove that quadrilateral $AKCB$ is cyclic and $2\alpha+2m+2n=\alpha+y+90-m-n+\alpha+m+2n$ $\implies$ we should prove:$90+y=2m+n$ which we proved. Done!