The problem is about geometric properties of a triangle ABC. 1) a) Construct points $M$ and $N$ such that $\vec{AM} = -\frac{2}{3}\vec{AB}$ and $\vec{NA} = \frac{2}{3}\vec{AC}$. b) Prove that lines $(MN)$ and $(BC)$ are parallel. c) Let $S$ and $T$ be the midpoints of $[BC]$ and $[MN]$ respectively. Prove that points $A$, $S$, and $T$ are collinear. 2) a) Construct points $I$ and $J$ such that $\vec{AI} = \frac{1}{3}\vec{AB}$ and $\vec{AJ} = 3\vec{AC}$. b) Prove that lines $(BJ)$ and $(IC)$ are parallel.

1. Problem Description

The problem is about geometric properties of a triangle ABC.

1) a) Construct points $M$ and $N$ such that $\vec{AM} = -\frac{2}{3}\vec{AB}$ and $\vec{NA} = \frac{2}{3}\vec{AC}$.

b) Prove that lines $(MN)$ and $(BC)$ are parallel.

c) Let $S$ and $T$ be the midpoints of $[BC]$ and $[MN]$ respectively. Prove that points $A$, $S$, and $T$ are collinear.

2) a) Construct points $I$ and $J$ such that $\vec{AI} = \frac{1}{3}\vec{AB}$ and $\vec{AJ} = 3\vec{AC}$.

b) Prove that lines $(BJ)$ and $(IC)$ are parallel.

2. Solution Steps

1) b)

We want to show that $(MN)$ and $(BC)$ are parallel. This means that the vectors $\vec{MN}$ and $\vec{BC}$ are collinear.

We have $\vec{MN} = \vec{MA} + \vec{AN} = -\vec{AM} - \vec{NA} = - (-\frac{2}{3}\vec{AB}) - (\frac{2}{3}\vec{AC}) = \frac{2}{3}\vec{AB} - \frac{2}{3}\vec{AC} = \frac{2}{3}(\vec{AB} - \vec{AC}) = \frac{2}{3}(\vec{AB} + \vec{CA}) = \frac{2}{3}\vec{CB} = -\frac{2}{3}\vec{BC}$.

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

1) c)

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

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

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

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

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

Since $\vec{AT}$ is a scalar multiple of $\vec{AS}$, the vectors $\vec{AT}$ and $\vec{AS}$ are collinear. This means that the points $A$, $S$, and $T$ are collinear.

2) b)

We want to show that $(BJ)$ and $(IC)$ are parallel, which means that $\vec{BJ}$ and $\vec{IC}$ are collinear.

$\vec{BJ} = \vec{BA} + \vec{AJ} = -\vec{AB} + 3\vec{AC}$.

$\vec{IC} = \vec{IA} + \vec{AC} = -\vec{AI} + \vec{AC} = -\frac{1}{3}\vec{AB} + \vec{AC}$.

$\vec{BJ} = 3(-\frac{1}{3}\vec{AB} + \vec{AC}) = 3\vec{IC}$.

Since $\vec{BJ} = 3\vec{IC}$, the vectors $\vec{BJ}$ and $\vec{IC}$ are collinear, so the lines $(BJ)$ and $(IC)$ are parallel.

3. Final Answer

1) b) $(MN)$ and $(BC)$ are parallel.

1) c) $A$, $S$, and $T$ are collinear.

2) b) $(BJ)$ and $(IC)$ are parallel.