Given: A is the midpoint of $\overline{\text{OX}}$;

$\overline{\text{AB}}\parallel \overline{\text{XY}}$; $\overline{\text{BC}}\parallel \overline{\text{YZ}}$

Prove: $\overline{\text{AC}}\parallel \overline{\text{XZ}}$

Here A is the midpoint of $\overline{OX}$, $\overline{AB}\parallel \overline{XY}$ and $\overline{BC}\parallel \overline{YZ}$.

In $\Delta XOY$, A is the midpoint of $\overline{OX}$,

And, $\overline{AB}\parallel \overline{XY}$.

So,

B is a midpoint of $\overline{OY}$.

(According to the converse of midpoint theorem, if a line is drawn through the midpoint of a side of a triangle parallel to the other side, then it bisects the third side.)

In $\Delta YOZ$, B is the midpoint of $\overline{OY}$,

Also, $\overline{BC}\parallel \overline{YZ}$.

So,

C is a midpoint of $\overline{OZ}$.

(According to the converse of midpoint theorem, if a line is drawn through the midpoint of a side of a triangle parallel to the other side, then it bisects the third side.)

In $\Delta XOZ$, A is a midpoint of $\overline{OX}$ and C is a midpoint of $\overline{OZ}$.

Also,

According to the midpoint theorem, the line joining the midpoints of two sides of a triangle is parallel to the third side of the triangle.

So,

$\overline{AC}\parallel \overline{XZ}$

Therefore, $\overline{AC}\parallel \overline{XZ}$(proved).