Let $ABC$ be a triangle. $B'$ and $C'$ are the midpoints of segments $[AC]$ and $[AB]$ respectively. $k$ is a real number. Points $D$ and $E$ are defined by $\vec{AD} = k \vec{AB}$ and $\vec{CE} = k \vec{CA}$. $I$ is the midpoint of $[DE]$. Prove that $B'$, $C'$, and $I$ are collinear.

1. Problem Description

Let $ABC$ be a triangle. $B'$ and $C'$ are the midpoints of segments $[AC]$ and $[AB]$ respectively. $k$ is a real number. Points $D$ and $E$ are defined by $\vec{AD} = k \vec{AB}$ and $\vec{CE} = k \vec{CA}$. $I$ is the midpoint of $[DE]$. Prove that $B'$, $C'$, and $I$ are collinear.

2. Solution Steps

We want to show that $\vec{C'I} = \lambda \vec{C'B'}$ for some real number $\lambda$.

Since $I$ is the midpoint of $[DE]$, we have

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

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

Therefore,

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

Since $C'$ and $B'$ are midpoints of $AB$ and $AC$ respectively, we have

$\vec{AC'} = \frac{1}{2}\vec{AB}$ and $\vec{AB'} = \frac{1}{2}\vec{AC}$

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

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

Thus, $\vec{C'I} = (1-k)\vec{C'B'}$. Since $\vec{C'I}$ is a scalar multiple of $\vec{C'B'}$, the points $B'$, $C'$ and $I$ are collinear.

3. Final Answer

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