The problem provides a triangle $ABC$, a point $P$ on segment $[AB]$, and a line through $P$ parallel to $BC$ that intersects $AC$ at $Q$. Points $I$ and $J$ are the midpoints of segments $[AP]$ and $[AB]$ respectively. The goal is to show that lines $IQ$ and $JC$ are parallel.

1. Problem Description

The problem provides a triangle $ABC$, a point $P$ on segment $[AB]$, and a line through $P$ parallel to $BC$ that intersects $AC$ at $Q$. Points $I$ and $J$ are the midpoints of segments $[AP]$ and $[AB]$ respectively. The goal is to show that lines $IQ$ and $JC$ are parallel.

2. Solution Steps

Let $A$ be the origin, i.e. $A = (0, 0)$. Let $\vec{b}$ be the vector from $A$ to $B$ and $\vec{c}$ be the vector from $A$ to $C$.

Since $P$ lies on the line segment $AB$, we can write $\vec{p} = \lambda \vec{b}$ for some scalar $\lambda$ such that $0 < \lambda < 1$.

Since $PQ$ is parallel to $BC$, the vector $\vec{q} - \vec{p}$ is parallel to $\vec{c} - \vec{b}$. Thus, $\vec{q} - \vec{p} = \mu(\vec{c} - \vec{b})$ for some scalar $\mu$.

Then $\vec{q} = \vec{p} + \mu(\vec{c} - \vec{b}) = \lambda \vec{b} + \mu(\vec{c} - \vec{b})$. Also, $Q$ lies on the line segment $AC$, so $\vec{q} = \alpha \vec{c}$ for some scalar $\alpha$.

Thus, $\alpha \vec{c} = \lambda \vec{b} + \mu(\vec{c} - \vec{b}) = (\lambda - \mu)\vec{b} + \mu \vec{c}$. Since $\vec{b}$ and $\vec{c}$ are linearly independent, we must have $\lambda - \mu = 0$, so $\lambda = \mu$. Thus, $\alpha = \lambda$, and $\vec{q} = \lambda \vec{c}$.

Since $I$ is the midpoint of $AP$, $\vec{i} = \frac{1}{2} \vec{p} = \frac{1}{2} (\lambda \vec{b}) = \frac{\lambda}{2} \vec{b}$.

Since $J$ is the midpoint of $AB$, $\vec{j} = \frac{1}{2} \vec{b}$.

The vector $\vec{iq} = \vec{q} - \vec{i} = \lambda \vec{c} - \frac{\lambda}{2} \vec{b} = \lambda (\vec{c} - \frac{1}{2} \vec{b})$.

The vector $\vec{jc} = \vec{c} - \vec{j} = \vec{c} - \frac{1}{2} \vec{b}$.

Then $\vec{iq} = \lambda (\vec{c} - \frac{1}{2} \vec{b}) = \lambda \vec{jc}$. Since $\vec{iq}$ is a scalar multiple of $\vec{jc}$, $IQ$ is parallel to $JC$.

3. Final Answer

The two straight lines $(IQ)$ and $(JC)$ are parallel.