How to join two points with only a small ruler. Given two points $A $ and $B $ in the  plane .
is it possible to draw the segment $[A,B] $ with only a ruler whose length is much smaller the the distance $\| \vec {AB }\|.$
I know the answer is yes and it uses the harmonic conjugate notion, but after many attempts, i still don't see how it is possible.
Thanks a lot for any help.
 A: This is a classic problem in projective geometry. You want to apply the dual of Desargues' Theorem. This states the following: Suppose $\triangle PQR$ and $\triangle P'Q'R'$ are triangles in the projective plane (but the Euclidean plane will do). Suppose the three points $D = \overleftrightarrow{PQ}\cap\overleftrightarrow{P'Q'}$, $E=\overleftrightarrow{QR}\cap\overleftrightarrow{Q'R'}$, and $F=\overleftrightarrow{PR}\cap\overleftrightarrow{P'R'}$ are collinear. Then the three lines $\overleftrightarrow{PP'}$, $\overleftrightarrow{QQ'}$, and $\overleftrightarrow{RR'}$ are concurrent (for which one option is all parallel lines in the Euclidean plane).
As a hint to start, draw any line segments starting at $A$ (making a relatively small angle) and use your little ruler to extend both of these until they get somewhat near $B$. By applying this theorem, you should be able to draw a line segment starting at $B$ which will, when extended, pass through $A$.
A: Given $A,B$ you can get the midpoint of the segment between them just using a compass.  Swing two arcs with radii shorter than $AB$ and you will have two closer points with the same midpoint.  Keep going until you get a segment shorter than your ruler.  Now that you have the midpoint, the problem is reduced in half.  Keep going until the problem is reduced below the length of your ruler.  Join the right points.
