Let $ABC$ be a triangle. Let $B'$ and $C'$ be the midpoints of the segments $[AC]$ and $[AB]$, respectively. Let $k$ be a real number. Let $D$ and $E$ be points in the plane such that $\vec{AD} = k \vec{AB}$ and $\vec{CE} = k \vec{CA}$. Let $I$ be the midpoint of $[DE]$. We need to prove that $B'$, $C'$ and $I$ are collinear.

1. Problem Description

Let $ABC$ be a triangle. Let $B'$ and $C'$ be the midpoints of the segments $[AC]$ and $[AB]$, respectively. Let $k$ be a real number. Let $D$ and $E$ be points in the plane such that $\vec{AD} = k \vec{AB}$ and $\vec{CE} = k \vec{CA}$. Let $I$ be the midpoint of $[DE]$. We need to prove that $B'$, $C'$ and $I$ are collinear.

2. Solution Steps

We are given that $B'$ is the midpoint of $AC$, so $\vec{AB'} = \frac{1}{2} \vec{AC}$. Similarly, $C'$ is the midpoint of $AB$, so $\vec{AC'} = \frac{1}{2} \vec{AB}$.

Also, $I$ is the midpoint of $DE$, which implies

$\vec{AI} = \frac{1}{2} (\vec{AD} + \vec{AE})$.

We have $\vec{AD} = k \vec{AB}$.

We also have $\vec{CE} = k \vec{CA}$. So, $\vec{AE} = \vec{AC} + \vec{CE} = \vec{AC} + k \vec{CA} = \vec{AC} - k \vec{AC} = (1-k)\vec{AC}$.

Thus,

$\vec{AI} = \frac{1}{2} (k \vec{AB} + (1-k)\vec{AC})$.

We want to show that $B'$, $C'$, and $I$ are collinear. We can show this by proving that $\vec{B'I}$ and $\vec{B'C'}$ are collinear.

$\vec{B'I} = \vec{AI} - \vec{AB'} = \frac{1}{2} (k \vec{AB} + (1-k)\vec{AC}) - \frac{1}{2} \vec{AC} = \frac{1}{2} k \vec{AB} + \frac{1}{2} \vec{AC} - \frac{1}{2} k \vec{AC} - \frac{1}{2} \vec{AC} = \frac{1}{2} k (\vec{AB} - \vec{AC}) = \frac{1}{2} k \vec{CB}$.

$\vec{B'C'} = \vec{AC'} - \vec{AB'} = \frac{1}{2} \vec{AB} - \frac{1}{2} \vec{AC} = \frac{1}{2} (\vec{AB} - \vec{AC}) = \frac{1}{2} \vec{CB}$.

$\vec{B'I} = k \vec{B'C'}$.

Thus, $\vec{B'I}$ and $\vec{B'C'}$ are collinear.

Therefore, $B'$, $C'$, and $I$ are collinear.

3. Final Answer

$B', C', I$ are collinear.