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

1. Problem Description

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

2. Solution Steps

We will express all the vectors in terms of $\vec{AB}$ and $\vec{AC}$.

Since $C'$ is the midpoint of $[AB]$, we have $\vec{AC'} = \frac{1}{2} \vec{AB}$.

Since $B'$ is the midpoint of $[AC]$, we have $\vec{AB'} = \frac{1}{2} \vec{AC}$.

We are given $\vec{AD} = k \vec{AB}$.

Also, we are given $\vec{CE} = k \vec{CA}$. So, $\vec{AE} = \vec{AC} + \vec{CE} = \vec{AC} + k \vec{CA} = (1 - k) \vec{AC}$.

Since $I$ is the midpoint of $[DE]$, we have $\vec{AI} = \frac{1}{2} (\vec{AD} + \vec{AE})$.

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

Now, we want to find if $\vec{C'I}$ and $\vec{C'B'}$ are collinear.

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

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

We observe that $\vec{C'I} = (1 - k) \vec{C'B'}$.

Since $\vec{C'I}$ is a scalar multiple of $\vec{C'B'}$, the vectors $\vec{C'I}$ and $\vec{C'B'}$ are collinear.

Therefore, the points $C'$, $I$, and $B'$ are collinear (aligned).

3. Final Answer

$\vec{C'I} = (1 - k) \vec{C'B'}$