Fit a equilateral triangle on three arbitrary parallel lines with an edge and compass How can you fit a equilateral triangle on three arbitrary parallel lines with an edge and compass?
Well, I'm late to this, but I've been obsessing over this far too much to back out now.  So: Label your parallel lines $a$, $b$, and $c$ from bottom to top.  Construct a line $d$ with a "positive slope" that crosses $b$ such that the top right angle of their intersection is $60^\circ$.  Extend the line so that is crosses $c$.  Call the intersection of $d$ and $c$ point $A$.  Construct another line $e$ such that it lies to the left of $d$, is parallel to $d$, and is the same distance from $d$ as $a$ is from $b$.  Call the intersection of $e$ and $b$ point $B$.  Draw an arc centered at $B$ and having radius $AB$, and crossing $a$.  Call the intersection of the arc with $a$ point $C$.  $\triangle ABC$ is your triangle.



Why does this work?  Here's a hint: Start with an equilateral triangle $\triangle ABC$.  Draw any line through $A$, and then draw a parallel line through $B$.  Now identify the center $O$ of $\triangle ABC$.  You have now have a triangle and two lines.  Rotate the entire configuration $120^\circ$ around $O$.  What do you get?  Rotate another $120^\circ$.  What do you get now?
Given 3 parallel lines A B C (top to bottom), construct a large equilateral triangle between lines A and C where the bottom triangle leg is on line C and the top triangle vertex is on line A. Mark the bottom right point of the triangle point c1.  Also mark the intersecting points of lines B and the left leg of the triangle point as b1.  Set the compass point on point c1 and the compass lead on point b1.  Now draw an arc that intersects line A, at new point a1. Use the straight edge and draw the new equilateral triangle between points b1, c1, and a1.

I figured this out in 1965 and made a proof to confirm validity, but have long forgotten the proof.

Regards, Ray McKaig