Let $ABC$ be any triangle and let $P$ be any point of the segment $AB$. The parallel to the line $BC$ passing through $P$ meets the line $AC$ at $Q$. Let $I$ and $J$ be the midpoints of $AP$ and $AB$ respectively. We need to show that the two lines $IQ$ and $JC$ are parallel.

1. Problem Description

Let $ABC$ be any triangle and let $P$ be any point of the segment $AB$. The parallel to the line $BC$ passing through $P$ meets the line $AC$ at $Q$. Let $I$ and $J$ be the midpoints of $AP$ and $AB$ respectively. We need to show that the two lines $IQ$ and $JC$ are parallel.

2. Solution Steps

Let $M$ be the midpoint of the segment $AQ$.

Since $I$ is the midpoint of $AP$, the segment $IM$ is parallel to $PQ$ and $IM = \frac{1}{2} PQ$. This is a consequence of the midpoint theorem. Also, since $PQ$ is parallel to $BC$, then $IM$ is parallel to $BC$.

Since $M$ is the midpoint of $AQ$, we have $AM = \frac{1}{2} AQ$.

Also, since $J$ is the midpoint of $AB$, we have $AJ = \frac{1}{2} AB$.

Consider the triangle $ABQ$. Since $J$ is the midpoint of $AB$ and $M$ is the midpoint of $AQ$, then $JM$ is parallel to $BQ$ and $JM = \frac{1}{2} BQ$.

Consider the triangle $ABC$. Since $PQ$ is parallel to $BC$, we have that $\frac{AP}{AB} = \frac{AQ}{AC}$.

Since $I$ is the midpoint of $AP$, $AP = 2 AI$. Since $J$ is the midpoint of $AB$, $AB = 2 AJ$.

Then $\frac{2 AI}{2 AJ} = \frac{AQ}{AC}$, so $\frac{AI}{AJ} = \frac{AQ}{AC}$.

Consider the triangle $ACJ$ and the line $IQ$. If $\frac{AI}{AJ} = \frac{AQ}{AC}$, then $IQ$ is parallel to $JC$.

Since $I$ is the midpoint of $AP$ and $J$ is the midpoint of $AB$, we have $AI = \frac{1}{2} AP$ and $AJ = \frac{1}{2} AB$.

Thus, $\frac{AI}{AJ} = \frac{\frac{1}{2}AP}{\frac{1}{2}AB} = \frac{AP}{AB}$.

We are given that $PQ$ is parallel to $BC$, so by similar triangles, $\frac{AP}{AB} = \frac{AQ}{AC}$.

Therefore, $\frac{AI}{AJ} = \frac{AQ}{AC}$.

By the converse of Thales' theorem, $IQ$ is parallel to $JC$.

3. Final Answer

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