# Difference between revisions of "2006 Romanian NMO Problems/Grade 7/Problem 1"

## Problem

Let $ABC$ be a triangle and the points $M$ and $N$ on the sides $AB$ respectively $BC$, such that $2 \cdot \frac{CN}{BC} = \frac{AM}{AB}$. Let $P$ be a point on the line $AC$. Prove that the lines $MN$ and $NP$ are perpendicular if and only if $PN$ is the interior angle bisector of $\angle MPC$.

## Solution

Let $L$ be a point on $BC$ such that $N$ is the midpoint of $LC$, then $2CN$= $LC$, the given information is the same as \frac{LN}{BC} = \frac{AM}{AB} $, applicating Thales theorem it follows that$ML $is parallel to$AC$. Let$ (Error compiling LaTeX. ! Missing $inserted.)R $be the point on$MN $such that$MN $=$NR $, in view of$MN $=$NR $and$LN $=$NC $it follows that$RLMC $is a parallelogram, implying that$CR $is parallel to$ML $, but we know that$ML $is parallel to$AC $, then$A $,$C $,$R $are collineal.$MN $is perpendicular to$PN $if and only if$NP $is the perpendicular bisector of$MC $if and only if$PN $is the angle bisector of$\angle MPR $if and only if$PN $is the angle bisector of$\angle MPC$, as requiered.