Given triangle $ABC$, we construct points $M$ and $N$ such that $\vec{AM} = -\frac{2}{3}\vec{AB}$ and $\vec{AN} = -\frac{2}{3}\vec{AC}$. We need to prove that lines $(MN)$ and $(BC)$ are parallel. Then, we let $S$ and $T$ be the midpoints of segments $[BC]$ and $[MN]$ respectively. We need to prove that points $A$, $S$, and $T$ are collinear.

1. Problem Description

Given triangle $ABC$, we construct points $M$ and $N$ such that $\vec{AM} = -\frac{2}{3}\vec{AB}$ and $\vec{AN} = -\frac{2}{3}\vec{AC}$.

We need to prove that lines $(MN)$ and $(BC)$ are parallel.

Then, we let $S$ and $T$ be the midpoints of segments $[BC]$ and $[MN]$ respectively. We need to prove that points $A$, $S$, and $T$ are collinear.

2. Solution Steps

First, let's prove that $(MN)$ and $(BC)$ are parallel.

We have $\vec{AM} = -\frac{2}{3}\vec{AB}$ and $\vec{AN} = -\frac{2}{3}\vec{AC}$.

Then,

$\vec{MN} = \vec{AN} - \vec{AM} = -\frac{2}{3}\vec{AC} - (-\frac{2}{3}\vec{AB}) = -\frac{2}{3}\vec{AC} + \frac{2}{3}\vec{AB} = \frac{2}{3}(\vec{AB} - \vec{AC}) = -\frac{2}{3}(\vec{AC} - \vec{AB}) = -\frac{2}{3}\vec{BC}$.

Since $\vec{MN} = -\frac{2}{3}\vec{BC}$, the vectors $\vec{MN}$ and $\vec{BC}$ are collinear, thus the lines $(MN)$ and $(BC)$ are parallel.

Next, let's prove that $A$, $S$, and $T$ are collinear.

$S$ is the midpoint of $BC$, so $\vec{AS} = \frac{1}{2}(\vec{AB} + \vec{AC})$.

$T$ is the midpoint of $MN$, so $\vec{AT} = \frac{1}{2}(\vec{AM} + \vec{AN}) = \frac{1}{2}(-\frac{2}{3}\vec{AB} - \frac{2}{3}\vec{AC}) = -\frac{1}{3}(\vec{AB} + \vec{AC})$.

Then, $\vec{AT} = -\frac{2}{3} \cdot \frac{1}{2} (\vec{AB} + \vec{AC}) = -\frac{2}{3} \vec{AS}$.

Since $\vec{AT} = -\frac{2}{3}\vec{AS}$, the vectors $\vec{AT}$ and $\vec{AS}$ are collinear, which means that points $A$, $S$, and $T$ are collinear.

3. Final Answer

$\vec{AT} = -\frac{2}{3}\vec{AS}$