Let $ABC$ be a triangle and $A', B', C'$ the points in which its incircle touches the sides $BC, CA, AB$, respectively. We denote by $I$ the incenter and by $P$ its projection onto $AA' $. Let $M$ be the midpoint of the line segment $[A'B']$ and $N$ be the intersection point of the lines $MP$ and $AC$. Prove that $A'N $is parallel to $B'C'$
Problem
Source: 2012 Romania JBMO TST1 P5
Tags: geometry, incenter, incircle, parallel
Steve12345
08.10.2020 17:04
Easy if you know symmedian stuff. But I guess that wasn't the point on a junior comp. Obviously we introduce the intersection $AA'$ with the incircle. Let it be $S$. We have $SB'||PN$ so we have by Thales some ratios. Working with PoP we get the quad $PA'B'N$ is cyclic and obviously $AC'B'IP$ is cyclic. With this info we can bash angles and get that $A'N$ is parallel to $B'C'$.
DensSv
03.01.2025 16:05
We redefine $N'$ as the parallel from $A'$ to $B'C'$. Let $L=AI\cap A'N$, first of all we need to observe the fact that $AL\perp A'N'$, since $AI\perp B'C'$.
We start by observing that, by the reciprocal of Menelaus, we need to demonstrate that :
$$\frac{A'P}{AP}\cdot\frac{AN'}{B'N'}\cdot \frac{B'M}{MA'}=\frac{A'P}{AP}\cdot\frac{AN'}{B'N'}=1$$Which is equivalent to
$$\frac{A'P}{AP}\stackrel{?}{=}\frac{B'N'}{AN'} \Leftrightarrow \frac{AP}{AN'}\stackrel{?}{=}\frac{A'P}{B'N'}$$.
In order to demonstrate that, we need two claims
Claim 1 $\triangle AB'P\sim\triangle AN'A'$.
Notice that $APIB'$ is an inscribed quadrilateral since $\angle IB'P=\angle IPA=90^{\circ}$. So $\angle PB'A=\angle AIP=90^{\circ}-\angle IAP=\angle AA'N' $. Hence, by $AA$ $\triangle AB'P\sim\triangle AN'A'$.
With this claim we obtain the fact that $\frac{AP}{AN'}=\frac{AB'}{AA'}$.
Claim 2 $\frac{AB'}{AA'}=\frac{A'P}{B'N'}$.
We proved earlier that $\angle PB'A=\angle PA'N$, hence $PB'N'A'$ is an inscribed quadrilateral.
Now, $\triangle PMB'\sim\triangle A'MN$ hence $\frac{B'M}{MN}=\frac{B'P}{A'N}=\frac{AB'}{AA'}$. $(1)$
Also, $\triangle PMA'\sim\triangle B'MN$ so $\frac{A'P}{B'N}=\frac{A'M}{MN}=\frac{B'M}{MN}$.$ (2)$
From $(1)+(2)$ we conclude the claim.
From Claim 1 +Claim 2 we obtain that $\frac{AP}{AN'}=\frac{A'P}{B'N'}$.
As stated above, by the reciprocal of Menelaus theorem we obtain the fact that $N'=N$, done.