Would it help if we replaced "line segment" with "chord of a circle with missing center"?

Note: If the circle had a center, then the Poncelet-Steiner theorem would guarantee a construction.

Regards,
Bractals

According to a Russian book by Zeitel Geometry of Straightedge and Geometry Compass, the center of a circle could be found in the presence of additional shapes, in particular if there is a parallelogram. However, if there is a parallelogram, you do not need a circle to find the midpoint of a segment.

>draw two chords parallel to the given lines

Could you expand on this construction.

>find their midpoints

That was the problem.

Given two parallel lines you can draw a line parallel to the two through any point.

Let P be the point. Draw a transversal through P to intersect the two lines in A and B, say. Let Q be a point on the transversal different from all the above. Draw a transversal through Q to intersect the two lines in C and D, respectively.

Let M = AD /\ BC,
Let N = QM /\ CP,
Let S = AN /\ CQ.

Then PS||BD.

You can perform this twice by choosing two points on the circle to get two parallel chords.

>> find their midpoints
> That was the problem.

No, the problem was to find the midpoint of a segment which apparently did not belong to either of the two parallel lines.

If the segment lies on one of the parallel lines (as is in the question of dividing in half two parallel chords) the construction is easy and is the most basic construction in the geometry of straightedge. (Repeat the construction above with Q = P. Then MP divides both AC and BD in half.)