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?

For brevity, I will not go into details about how to construct perpendiculars and 30-degree angles using straightedge and compass. Same for doubling lengths. Translating those operations to compass-and-straightedge primitives is left to the reader. Given the three parallel lines and a vertex C arbitrarily fixed on one of those lines, I will find the two locations of another vertex, named E and E', that are suited for the fit, and I will not lose words about the rest because that should be obvious.

In C, construct a line perpendicular to the three parallel lines, and name the new intersection points A and B. Now have a look at the attached figure, which is symmetric with respect to the perpendicular line AC, and assume that all filled triangles are equilateral. In this demonstration, |AC| >= |AB| >= |BC| for tidyness, but it turns out that algebraically there is no such restriction for the scheme to work.

Rotate the kite CEFE' 60 degress around E to find that it is congruent to DEE'G. Therefore |DG|=|DE|=|DC|, so |CG|=2|CB|. Using this you can locate G and then branch off CG in G at angles of +/-30 degrees, intersecting with the parallel line through A to find E resp. E'.

I've found a different invariant that solves this problem: two circles bisecting a perpendicular between any two parallels cut each other at points that are exact middles of the side opposite to the vertex on the third parallel.

If you want more explanations and sketches please go to:

http://romanyandronov.elementfx.com/pse/ryapserac03.html

Sorry for providing the link - it's a small part of a bigger thing. In my articles I'm interested not in the answers themselves but rather in the ways one can find them. You may find other things of interest there.

This is an eleven-step construction, but I think that number can be reduced.

Outline: random point on any parallel, perpendicular through it, two circles bisecting any segment between any two lines, line through their intersection point and a point on the remaining parallel, perpendicular to that last line locates two remaining vertexes: