dividing a segment into n equal parts
Splitting a segment in two is easy — bisect it. But suppose you need to cut AB into 5 equal pieces, or 7, or any number n, without ever measuring its length. Repeated bisection only gives you halves, quarters, eighths; it can never produce fifths. The clever fix uses parallels instead of arcs.
Draw any ray from A at a convenient angle, going off to the side. Open the compass to ANY fixed width and step off n equal marks along that ray, calling them P_1, P_2, ..., P_n; these are equally spaced because each hop is the same compass width, even though their actual length is irrelevant. Join the last mark P_n to the endpoint B. Now, through each of P_1, ..., P_(n-1), construct a line parallel to P_n B. By the side-splitter theorem, those parallels cut AB into n equal parts. The trick is that equal spacing on the slanted ray transfers, via parallels, to equal spacing on AB.
This is a beautiful application of similar triangles: the parallel lines create nested similar triangles, and proportional sides do the dividing for you. It works for any n at all, which is striking — yet it does NOT contradict the impossibility results. Dividing a SEGMENT into n equal lengths is always possible; dividing an ANGLE into n equal parts is a different, far harder problem that fails already at n = 3 for a general angle.
To cut AB into 5 parts: from A draw a slanted ray and step off 5 equal compass-widths to P_1..P_5. Join P_5 to B, then draw lines through P_1..P_4 parallel to P_5 B; they slice AB into 5 equal pieces.
Equal hops on a slanted ray, carried back by parallels, divide AB evenly — similar triangles in action.
Dividing a segment into n equal lengths is always possible for every n. Do not confuse it with dividing an angle into n equal parts, which is generally impossible — trisecting an arbitrary angle already fails.