Given a triangle $ABC$, points $M$ and $N$ are defined such that $\vec{AM} = -\frac{2}{3}\vec{AB}$ and $\vec{AN} = -\frac{2}{3}\vec{AC}$. (1) We need to show that lines $(MN)$ and $(BC)$ are parallel. (2) Let $S$ and $T$ be the midpoints of segments $[BC]$ and $[MN]$, respectively. We need to show that points $A$, $S$, and $T$ are collinear.

1. Problem Description

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

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

(2) Let $S$ and $T$ be the midpoints of segments $[BC]$ and $[MN]$, respectively. We need to show that points $A$, $S$, and $T$ are collinear.

2. Solution Steps

(1) To show that $(MN)$ and $(BC)$ are parallel, we need to show that $\vec{MN}$ and $\vec{BC}$ are collinear.

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

Since $\vec{MN} = -\frac{2}{3}\vec{BC}$, $\vec{MN}$ and $\vec{BC}$ are collinear, which means that lines $(MN)$ and $(BC)$ are parallel.

(2) To show that $A$, $S$, and $T$ are collinear, we need to show that $\vec{AT}$ and $\vec{AS}$ are collinear.

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

Since $T$ is the midpoint of $MN$, $\vec{AT} = \frac{1}{2}(\vec{AM} + \vec{AN})$.

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

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

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

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

3. Final Answer

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