Alternative (and possibly simpler): Let two lines PQ and MM parallel to each other be given, and a point O contained between the region of the plane bounded by PQ and MM also be given. Let the segment to be constructed, a, be given.

(1) If a is less than the distance from PQ to MM (as defined by part 85), then there is no solution because any perpendicular dropped from PQ to MM (or vice-versa) would be smaller than any slant (part 51), and a will only form a part of the perpendicular.

(2) If a is not less than the distance from PQ to MM, then it can be constructed through O, with endpoints on PQ and MM, respectively, using the following process:

Choose a point A along MN, and setting the compass to a step equal to a, place the pin leg at A. Using the compass' pencil leg, describe an arc of radius a, and mark its intersection with PQ as B.

Draw a line from A to B. If O lies along A and B, then we are done, so suppose it does not. Then, through the point O, construct a line parallel to AB, and mark its intersections with PQ and MN as C and D, respectively.

ABCD is a parallelogram (because by hypothesis, PQ || MN, and this implies CB || DA as segments of respectively parallel lines, and CD || AB by construction), so by part 85, CD = AB = *a* as opposite sides of a parallelogram.

But O belongs to CD by construction, and so we have constructed the required segment.