Cristea's theorem: $ D,E,F$ are three points on the sides $ CB,BA,AC$ of triangle $ \triangle ABC$ and $ M \in AD.$ Then the transversal $ EF$ goes through $ M$ if and only if the following relation holds: $ \overline{DC} \cdot \frac {\overline{EB}}{\overline{EA}} + \overline{BD} \cdot \frac {\overline{FC}}{\overline{FA}} = \overline{BC} \cdot \frac {\overline{MD}}{\overline{MA}}$ Particularly, $ EF$ passes through the centroid of $ \triangle ABC \Longleftrightarrow \frac {\overline{EB}}{\overline{EA}} + \frac {\overline{FC}}{\overline{FA}} = 1.$

Let $\frac{BM}{MA}=x$. Then $M=\frac{x}{x+1}A+\frac{1}{x+1}B$ $\frac{CN}{NA}=1-x$, so $N=\frac{1-x}{2-x}A+\frac{1}{2-x}C$ $\frac{x+1}{3}M+\frac{2-x}{3}N=\frac{1}{3}A+\frac{1}{3}B+\frac{1}{3}C$, which is the centroid. Since $\frac{x+1}{3}+\frac{2-x}{3}=1$, and it is easy to see that neither of these are negative, we therefore have that the centroid lies on the segment MN. Cheers, Rofler

It's POL 6 in 1969 IMO Longlist

An old-fashioned proof: Let $MN$ intersect $BC$ at $P$ and median $AD$ at $X$. Applying Menelaus to triangles $ACD$ and $ABD$ with the transversal $NXP$ and $XMP$ respectively we get $\frac{CN}{AN}=\frac{CP}{PD} \cdot \frac{DX}{AX}\; (\; 1 \;)$ and $\frac{BM}{MA}=\frac{PB}{PD}\cdot\frac{DX}{AX}\; (\; 2 \;)$; adding side by side the two equalities we get $\frac{BM}{AM}+\frac{CN}{AN}=\frac{DX}{AX}\cdot \left(\frac{PB+PC}{PD}\right) \; (\; 3 \; )$, but from the problem we know $\frac{BM}{AM}+\frac{CN}{AN}=1$ and, from $D$ being midpoint of $[BC]$ we get $PB+PC=2\cdot PD$ and, substituting in $(3)$ we get $AX=2\cdot DX$, i.e. $X\equiv G$, done. Best regards, sunken rock

Let $\frac{BM}{MA}=\frac{s}{1-s},\ \frac{CN}{NA}=\frac{t}{1-t}\ (0<s,\ t<1)$, for the centroid $G$, $\overrightarrow{AG}=\frac 13(\overrightarrow{AB}+\overrightarrow{AC})=\frac 13\left(\frac{1}{1-s}\overrightarrow{AM}+\frac{1}{1-t}\overrightarrow{AN}\right)$ $\therefore \frac{BM}{MA}+\frac{CN}{NA}=1\Longleftrightarrow \frac{s}{1-s}+\frac{t}{1-t}=1$ $\Longleftrightarrow -1+\frac{1}{1-s}-1+\frac{1}{1-t}=1$ $\Longleftrightarrow \frac 13\cdot \frac{1}{1-s}+\frac 13\cdot \frac{1}{1-t}=1$, which gives the collinearity of $M, G , N.$ $Q.E.D.$

Dear Mathlinkers, here Problem 6 Sincerely Jean-Louis