53. An remarkable concurrency.

by Virgil Nicula, Jul 9, 2010, 10:15 PM

Quote:
Problem. Given a triangle $ABC$ and the midpoints $M$ , $N$ , $P$ of its sides $[BC]$ , $[CA]$ , $[AB]$ . Let $X$ be a point in the plane of triangle $ABC$ and let the lines $AX$ , $BX$ , $CX$ intersect the lines $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ respectively. Denote by $D'$ , $E'$ , $F'$ the midpoints of the segments $AD$ , $BE$ , $CF$ . Prove that the lines $MD'$ , $NE'$ and $PF'$ concur at one point.

Proof. Since $N$ and $D'$ are the midpoints of $[CA]$ and $[AD]$ of $\triangle CAD$ , we must have $ND'\parallel CD$ . In other words, $ND'\parallel BC$ . Similarly, $PD'\parallel BC$ . Thus, $N$ , $D'$ and $P$ lie on one

and the same line parallel to $BC$ . In other words, $D'$ lies on $NP$ and we have $NP\parallel BC$ . Hence, by the Thales' theorem, $\frac{ND^{\prime}}{D^{\prime}P}=\frac{CD}{DB}$ . Similarly, $E'$ lies on $PM$ and satisfies

$\frac{PE^{\prime}}{E^{\prime}M}=\frac{AE}{EC}$ and $F'$ lies on $MN$ and satisfies $\frac{MF^{\prime}}{F^{\prime}N}=\frac{BF}{FA}$ . Thus, $\frac{ND^{\prime}}{D^{\prime}P}\cdot\frac{PE^{\prime}}{E^{\prime}M}\cdot\frac{MF^{\prime}}{F^{\prime}N}=$ $\frac{CD}{DB}\cdot\frac{AE}{EC}\cdot\frac{BF}{FA}=$ $\frac{CD}{DB}\cdot\frac{BF}{FA}\cdot\frac{AE}{EC}$ . But $AD$ , $BE$ , $CF$ concur (namely, at $X$).

Thus, by Ceva's theorem, $\frac{CD}{DB}\cdot\frac{BF}{FA}\cdot\frac{AE}{EC}=1$. Hence, $\frac{ND^{\prime}}{D^{\prime}P}\cdot\frac{PE^{\prime}}{E^{\prime}M}\cdot\frac{MF^{\prime}}{F^{\prime}N}=1$ . Thus, by Ceva's theorem, applied to $\triangle MNP$ and to $D'\in NP$ , $E'\in PM$ , $F'\in MN$ obtain

that $MD'$ , $NE'$ and $PF'$ must concur.

Virgil Nicula wrote:
Let $ABC$ be a triangle and the points $P(x, y, z),\ R(u, v, w)$ with barycentric coodonates relative to the triangle $ABC$.

I note $X,\ Y,\ Z$ which belong to $BC,\ CA,\ AB$ respectively such that $PX\parallel AR,\ PY\parallel BR,\ PZ\parallel CR$.

Prove that $P$ is the centroid of $\triangle XYZ$ iff there is the relations $\frac{u(v+w)}{x}=\frac{v(w+u)}{y}=\frac{w(u+v)}{z}$ .
This post has been edited 6 times. Last edited by Virgil Nicula, Nov 27, 2015, 7:27 AM

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