Two-sided ruler constructions 6: Parallel lines

This is the sixth in a series of blog posts about two-sided ruler constructions. Here are all the blog posts in the series:

1. Introduction
2. Fundamentals
3. Rhombuses
4. Copying and cutting
5. Perpendicular lines
6. Parallel lines (you are here)
7. Circles without circles
8. Equilateral triangle and regular pentagon

This blog post is about drawing lines parallel to other lines, but through specific points. One of the fundamental things the two-sided ruler can do is draw lines parallel to other lines that are exactly one ruler width away. But what if you want your parallel line to be more or less than that distance away? Well it can be done.

Near the end of the last blog post, I noted that the construction for drawing a line perpendicular to an existing line through a point off the line had a couple of parallel lines in its diagram, and I wondered if I could put that diagram in the correct spot to draw the parallel lines where I want them.

And it is possible. Here’s the construction I came up with that’s in the spirit of the perpendicular line construction.

Draw a line parallel to an existing line through a point not on that line (using a perpendicular line)
0. Start with a line and a point not on that line. Call the point \(Q\).
1. Align the point \(Q\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. Call by \(A\) the point where the line through \(Q\) meets the original line and by \(Y\) the point where the other parallel line meets the original line. Note it’s important to make sure \(Y\) and \(A\) are on opposite sides of the position of \(Q\) if you projected it down to the line.
2. Side-align the line you just drew through \(Y\) and draw along the other side of the ruler to make a third parallel line. You only need enough of this line to find where it meets the original line. Call this meeting point \(B\).
3. The points \(Y\) and \(B\) were already cross-aligned in step 2. Cross-align them in the other direction and draw along the side of the ruler through \(B\). Call by \(P\) the point where this line meets the line \(QA\).
4. Draw the line through \(P\) and \(Y\).
5. Draw the line through \(Q\) and \(B\) and find where it meets the line \(PY\). Call this point \(X\).
6. Draw the line through \(A\) and \(X\) and find where it meets the line \(BP\). Call this point \(S\).
7. Draw the line through \(Q\) and \(S\).
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\).
Video here

I have carefully made the labels of the points match up with the earlier construction, but the points and lines were drawn almost in the reverse order, so I think I need a new proof.

Proof:

Triangle \(\triangle APB\) was constructed by the same method for drawing a line perpendicular to a specific point on a line, and so, just as in that proof, it’s an isosceles triangle. In particular the sides \(AP\) and \(BP\) are equal, the angles \(\angle PAB\) and \(\angle PBA\) are equal, and the line \(PY\) meets \(AB\) in a right angle and bisects the angle \(APB\).

Consider the triangles \(\triangle APX\) and \(\triangle BPX\).

The sides \(AP\) and \(BP\) are equal, and the angles \(\angle APX\) and \(\angle BPX\) are equal, and the side \(PX\) is shared. Therefore \(\triangle APX\) and \(\triangle BPX\) are congruent.

Hence, \(\angle PAX\) and \(\angle PBX\) are equal.

Consider triangles \(\triangle APS\) and \(\triangle BPQ\).

The sides \(AP\) and \(BP\) are equal, the angles \(\angle PAX\) and \(\angle PBX\) are equal and the angles \(\angle APS\) and \(\angle BPQ\) are equal because they are shared. Therefore \(\triangle APS\) and \(\triangle BPQ\) are congruent.

Hence \(PS\) and \(PQ\) are the same length, making triangle \(QPS\) an isosceles triangle. The line \(PY\) bisects the angle \(\angle QPS\) and so it must meet the opposite side \(QS\) in a right angle.

Now \(QS\) and \(AB\) are both perpendicular to \(PY\) and are therefore parallel to each other.

End of proof!

Note I could have argued that last part differently by noticing that the ratio \(PQ:PA\) is the same as the ratio \(PS:PB\) and going via similar triangles. I didn’t, because I wanted to match the spirit of the previous construction.

However, there is a way to do almost this same construction without making a right angle anywhere, and indeed without making any isosceles triangles or bisecting any angles. You just need the base \(AB\) to be bisected and it will work. Indeed, when Wernick and Birrell describe this construction, they just say to arrange three equally spaced points on the line using the process for doubling a line longer than a ruler width, but no particular instructions of where to put them.

Their construction is indeed particularly useful if you do already happen to have three equally spaced points on that original line. My construction requires you to make the points yourself or the later lines won’t go through the right points, so theirs is very nice if you don’t want to make them yourself.

Draw a line parallel to an existing line through a point not on that line (using three existing equally spaced points on the line)
0. Start with a line and a point not on that line. Also start with three equally-spaced points on the line. Call the point not on the line \(Q\). Call the three points on the line \(A\), \(Y\) and \(B\) in order.
1. Draw the line through \(Q\) and \(A\).
2. Draw any line through \(B\) that meets \(QA\) in a point other than \(Q\) and \(A\). Call this point \(P\).
3. Draw the line through \(P\) and \(Y\).
4. Draw the line through \(Q\) and \(B\) and find where it meets the line \(PY\). Call this point \(X\).
5. Draw the line through \(A\) and \(X\) and find where it meets the line \(PB\). Call this point \(S\).
6. Draw the line through \(Q\) and \(S\).
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\).
Video here

It’s truly remarkable to me that this construction works. My favourite part is that it will work no matter which direction \(A\), \(Y\) and \(B\) are labelled and no matter where the point \(P\) is on the line \(QA\) . The diagrams produced will look similar from a distance, but will be labelled very differently. (Wernick and Birrell don’t draw all the other diagrams, but I wanted to be sure it really worked.)

The point \(A\) is closest to \(Q\). The point \(P\) is on the far side of \(Q\) from \(A\).

The point \(A\) is furthest from \(Q\). The point \(P\) is on the far side of \(Q\) from \(A\).

The point \(A\) is closest to \(Q\). The point \(P\) between \(Q\) and \(A\).

The point \(A\) is furthest from \(Q\). The point \(P\) is between \(Q\) and \(A\).

The point \(A\) is closest to \(Q\). The point \(P\) is on the far side of \(A\) from \(Q\).

The point \(A\) is futhest from \(Q\). The point \(P\) is on the far side of \(A\) from \(Q\).

Wenrick asserts that this construction works by citing the “harmonic properties of a complete quadrilateral” and Birrell properly proves it works using Ceva’s theorem, a very cool theorem that I learned because she used it. However, I’m pretty sure Ceva’s theorem won’t apply if the points \(P\) and \(X\) are on opposite sides of the original line, so I’ve just proved it directly. I did take inspiration from the proof of Ceva’s theorem, which uses areas.

Proof:

Consider triangles \(\triangle AYX\) and \(\triangle BYX\). If \(AY\) and \(BY\) are considered their bases, then note these bases are on the same line and the triangles also share the opposite vertex \(X\), so they have the same height. Since the bases and heights are equal, they must have the same area. I’ll use the modulus straight line brackets to mean the area of a triangle so I can write \(|\triangle AYX|=|\triangle BYX|\).

Now consider triangles \(\triangle AYP\) and \(\triangle BYP\). They also have aligned equal bases \(AY\) and \(BY\) and shared vertex \(P\), so \(|\triangle AYP|=|\triangle BYP|\).

Now consider triangles \(\triangle AXP\) and \(\triangle BXP\).
If \(X\) and \(P\) are on opposite sides of \(Y\), then \(\triangle AXP\) is made by joining \(\triangle AYX\) and \(\triangle AYP\) and \(\triangle BXP\) is made by joining \(\triangle BYX\) and \(\triangle BYP\).
If \(X\) and \(P\) are on the same side of \(Y\), then \(\triangle AXP\) is made by removing \(\triangle AYX\) from \(\triangle AYP\) and \(\triangle BXP\) is made by removing \(\triangle BYX\) from \(\triangle BYP\).
Either way, since the triangles involved in each calculation have the same areas as the matching triangles in the other calculation, the resulting triangles have the same area too.
That is, \(|\triangle AXP|=|\triangle BXP|\).

Now consider the two line segments \(QA\) and \(QP\) on the same line. Two triangles can be created with each of these segments as bases by choosing the opposite vertex to be \(X\) or \(B\).

The triangles \(\triangle QAX\) and \(\triangle QPX\) have aligned bases and shared vertex \(X\) so they have the same height. This means their areas are in the same proportion as \(QA\) and \(QP\). That is, \[\frac{QA}{QP}=\frac{|\triangle QAX|}{|\triangle QPX|}\].

Similarly, the triangles \(\triangle QAB\) and \(\triangle QPB\) have aligned bases and shared vertex \(B\) so they have the same height and their areas are in the same proportion as \(QA\) and \(QP\). That is, \[\frac{QA}{QP}=\frac{|\triangle QAB|}{|\triangle QPB|}\].

Now, if two fractions are equal, you can subtract (or add) the numerators and denominators and produce another equal fraction.

(A quick proof if you don’t believe it, because it does feel surprising…
Suppose \(\frac{a}{b} = \frac{c}{d}\).
Begin by multiplying both parts of these equations by both \(b\) and \(d\):
\[\begin{aligned}\frac{a}{b} &= \frac{c}{d} \\ ad &=bc \\ ad-ab &=bc -ab\\ a(d-b) &= (c-a)b \\ \frac{a}{b} &= \frac{c-a}{d-b} \end{aligned}\]
So there you go.)

Hence, \[\frac{QA}{QP}=\frac{|\triangle QAB| – |\triangle QAX|}{|\triangle QPB|-|\triangle QPX|}\].

Notice that removing \(\triangle QAX\) from \(\triangle QAB\) leaves \(\triangle BAX\) and removing \(\triangle QPX\) from \(\triangle QPB\) leaves triangle \(BPX\). Therefore \[\frac{QA}{QP}=\frac{|\triangle BAX|}{|\triangle BPX|}\].

Similar reasoning on the other side of the diagram concludes that \[\frac{SB}{SP}=\frac{|\triangle ABX|}{|\triangle APX|}\].

Note that \(\triangle BAX\) is the same as \(\triangle ABX\) and \(|\triangle APX| = |\triangle BPX|\), so \[\frac{SB}{SP}=\frac{|\triangle BAX|}{|\triangle BPX|}=\frac{QA}{QP}\]

Just a little more algebra to do:

\[\begin{aligned}\frac{SB}{SP}&=\frac{QA}{QP}\\ 1- \frac{SB}{SP}&=1- \frac{QA}{QP} \\ \frac{SP}{SP}- \frac{SB}{SP}&=\frac{QP}{QP}- \frac{QA}{QP} \\ \frac{SP-SB}{SP}&=\frac{QP-QA}{QP} \\ \frac{BP}{SP}&=\frac{AP}{QP} \end{aligned}\]

Consider triangles \(\triangle APB\) and \(\triangle QPS\). They share the angle at \(P\) and their matching sides on either side of that angle are in the same proportion. Therefore the triangles are similar.

Hence \(\angle PAB = \angle PQS\), which means lines \(AB\) and \(QS\) are parallel.

End proof!

This is an epic proof with a lot of ninja algebra, and ninja triangle-finding so I’m proud of figuring it out.

A very interesting thing about this construction is that after you’ve found the three equally-spaced points, you don’t use both of the sides of the ruler at once, so you could do it with a one-sided ruler. That’s pretty cool in its own way. Still, for some reason I find the construction fiddly. I can’t put my finger on why, but there it is.

I have discovered an alternative way to use three equally spaced points on the line to draw a parallel line. It’s a modification of Wenrick’s method of copying a line segment, and I think it’s rather cool.

Draw a line parallel to an existing line through a point not on that line (using three existing equally spaced points on the line, and an intermediate parallel line)
0. Start with a line and a point not on that line. Also start with three equally-spaced points on the line. Call the point not on the line \(Q\). Call the three points on the line \(W\), \(X\) and \(Y\) in order.
1. Side-align the original line and draw along the other side of the ruler to make a parallel line. It works best if this line is on the opposite side of the original line from \(Q\).
2. Draw the line through \(Q\) and \(W\) and find where it meets the parallel line you drew at step 1. Call this meeting point \(S\).
3. Draw the line through \(Q\) and \(X\) and ifind where it meets the parallel line you drew at step 1. Call this meeting point \(B\).
4. Draw the line through \(S\) and \(X\) .
5. Draw the line through \(B\) and \(Y\) and find where it meets the line \(SX\). Call this point \(A\).
6. Draw the line through \(Q\) and \(A\).
Done! The line \(QA\) passes through \(Q\) and is parallel to the original line \(WY\).
Video here

Proof:

Since \(WX\) is parallel to \(SB\) the triangles \(\triangle SQB\) and \(\triangle WQX\) are similar. Therefore the matching sides are in the same proportion. In particular \[\frac{QX}{QB}=\frac{WX}{SB}\]

Also the triangles \(\triangle SAB\) and \(\triangle XAY\) are similar, so that \[\frac{AY}{AB} =\frac{XY}{SB}\).

But the lengths of \(WX\) and \(XY\) are the same, so

\[\begin{align} \frac{QX}{QB} &= \frac{AY}{AB}\\ 1-\frac{QX}{QB} &= 1-\frac{AY}{AB} \\ \frac{QB}{QB}-\frac{QX}{QB} &= \frac{AB}{AB} \frac{AY}{AB} \\ \frac{QB-QX}{QB} &= \frac{AB-AY}{AB} \\ \frac{BX}{QB} &= \frac{BY}{AB}\end{align}\]

Now triangles \(\triangle QBA\) and \(\triangle XBY\) have a shared angle at \(B\) and the enclosing sides are in the same proportion, to they are similar triangles.

Therefore \(\angle BQA = \angle BXY\) and so \(QA\) must be parallel to \(XY\).

End proof!

I did say earlier that it works best if that intermediate parallel line is on the opposite side of the original line from \(Q\) but it does work even if it’s on the same side as \(Q\). I just find it a bit harder to draw. Here’s what it looks like in all three situations:

The intermediate parallel line is on the far side of the original line from \(Q\).

The intermediate parallel line is between the original line and \(Q\).

The intermediate parallel line is on the far side \(Q\) from the original line.

The proof above is designed for the case where that first parallel line is indeed on the opposite side of the original line from \(Q\), and the algebra in the middle needs to be modified to make it match the other cases.

I’m honestly surprised Wenrick didn’t include this method in his paper, since it is very similar to his method of copying a length. Me, I like it very much. There’s something cool to me about drawing an intermediate parallel line and somehow copying the parallelness to another line. Plus it uses the same number of lines as the previous construction (six, on top of any you used to make the three equally-spaced points).

Just like for Wenrick’s copy-a-length construction, it’s worth noting that this will work just fine with an intermediate parallel line that’s some other distance than one ruler-width away from the original line. So, if you happen to have such a line already, then this method is more efficient than the previous one. It’s only more efficient by one line, but that’s still better!

I do have one more way I’ve designed to construct a parallel line through a given point and it’s my favourite of all of them. I made it when I first looked at Wernick and Birrell’s construction, because it occurred to me that I have to draw three parallel lines to make those equally-spaced points and surely there must be a way to use those lines directly to construct the actual line I want. There is a way, and here it is.

Draw a line parallel to an existing line through a point not on that line (using three parallel lines)
0. Start with a line and a point not on that line. Call the point \(Q\).
1. Align the point \(Q\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. Call by \(A\) the point where the line through \(Q\) meets the original line and by \(Y\) the point where the other parallel line meets the original line.
2. Side-align the line you just drew through \(Y\) and draw along the other side of the ruler to make a third parallel line. Call by \(B\) the point where this line meets the original line.
3. Draw the line through \(Q\) and \(B\) and find where it meets the middle parallel line through \(Y\). Call this point \(X\).
4. Draw the line through \(A\) and \(X\) and find where it meets the third parallel line through \(B\). Call this point \(S\).
5. Draw the line through \(Q\) and \(S\).
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\).
Video here

Proof:

The three parallel lines through \(A\), \(Y\) and \(C\) all meet the original line at the same angle, so the lengths \(AY\) and \(YB\) are equal by the cross-align arcsine lemma.

Triangles \(\triangle AQB\) and \(\triangle YBX\) are similar because of the shared angle and the parallel lines. Since \(YB\) is half of \(AB\), this means \(YX\) is half as long as \(AQ\).

Triangles \(\triangle ABS\) and \(\triangle AYX\) are also similar, and \(AY\) is half as long as \(AB\), which means \(YX\) is also half as long as \(BS\).

Therefore \(AQ\) and \(BS\) are the same length.

Consider triangles \(\triangle AQB\) and \(\triangle SBQ\). Side \(BQ\) is shared, sides \(AQ\) and \(SB\) are the same length, and angles \(\angle AQB\) and \(\angle QBS\) are equal since they are alternate angles around a transversal to parallel lines.

There for \(\triangle AQB\) and \(\triangle SBQ\) are congruent, which means \(\angle ABQ = \angle BQS\). These are alternate angles around a transversal to \(AB\) and \(QS\).

Hence \(AB\) and \(QS\) are parallel.

End proof!

If this proof reminded you of the proof for Wernick’s method of copying a length, it’s not a coincidence, since the diagram is pretty much the same, only sideways, as it were. Indeed this is why I labelled the points the way I did for the intermediate parallel line way to make a parallel line, because that diagram too is pretty much this one but sideways.

(I actually didn’t read Wernick’s paper until long after I made this construction, and I noticed the connection when I came to write it all up.)

There’s another connection this construction and the other ones we’ve seen so far. It’s not just that I took the opportunity of the three parallel lines from Wernick and Birrell’s construction. If you take that point \(P\) from their construction and move it away from \(Q\) as far as you can, then the three lines through \(P\) stretch out further and further and become closer and closer to being parallel. This construction is the same construction but with the point \(P\) at infinity! This warms my finite geometer’s heart. I’m sure there’s some sort of projective geometry theorem that makes it all work (maybe those harmonic properties of the complete quadrilateral that Wernick mentioned), though I am very happy with my similar triangles and areas.

So, I’ve got four ways to draw a parallel line through a point not on the line.

The next thing I want to do is simultaneously draw a perpendicular line and a parallel line through a point not on the original line. This is a thing that’s useful in some later constructions I want to describe.

You could naively just draw a parallel line through the point and separately draw a perpendicular line through the point. The best versions of those constructions use six lines each, so I can definitely do it in twelve lines.

But actually if you already have one of the parallel or perpendicular lines, the other line can be drawn by going perpendicular to the line you just drew. That is, if you have a line parallel to another, then a perpendicular to that line is perpendicular to the original line; and if you have a line perpendicular to another, then a perpendicular to that line is parallel to the original.

And the construction for a perpendicular line through a point takes four lines. So the estimate is now six plus four (which is ten) lines. And it turns out either way I can reuse one of the lines from the first part when doing the second part, which gets me down to nine lines.

I can’t seem to make it happen in any less than nine, and honestly nine is pretty good considering that Wernick’s original method for drawing a parallel line through a point uses eight lines if you include the lines to make the equally spaced points. So here’s two constructions for drawing a perpendicular and parallel line at the same time, though I have to say I like the first one better.

Draw lines parallel then perpendicular to an existing line through a point not on that line
0. Start with a line and a point not on that line. Call the point \(Q\).
1. Align the point \(Q\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. You will need the line on the other side of the ruler from \(Q\) to be long enough to meet the perpendicular through \(Q\) once it’s drawn. Call by \(A\) the point where the line through \(Q\) meets the original line and by \(Y\) the point where the other parallel line meets the original line.
2. Side-align the line you just drew through \(Y\) and draw along the other side of the ruler to make a third parallel line. Call by \(B\) the point where this line meets the original line.
3. Draw the line through \(Q\) and \(B\) and find where it meets the middle parallel line through \(Y\). Call this point \(X\).
4. Draw the line through \(A\) and \(X\) and find where it meets the third parallel line through \(B\). Call this point \(S\).
5. Draw the line through \(Q\) and \(S\).
6. The line \(QA\) was side-aligned at step 2 to draw the line \(YX\). Side-align it on the other side and draw along the opposite edge of the ruler. You only need enough of this line to see where it meets \(QS\). Call this meeting point \(L\).
7. The points \(L\) and \(Q\) were cross-aligned at step 6. Cross-align them in the other direction and draw along the side of the ruler through \(L\). You only need enough of this line to see where it meets \(XY\). Call this meeting point \(M\).
8. Draw the line through \(M\) and \(Q\) long enough to meet the original line.
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\). The line \(MQ\) passes through \(Q\) and is perpendicular to the original line.
Video here

Draw lines perpendicular then parallel to an existing line through a point not on that line
0. Start with a line and a point not on that line. Call the point \(A\).
1. Align the point \(A\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. You will need the line on the other side of the ruler from \(A\) to be long enough to reach both the perpendicular and parallel lines through \(A\) once they are drawn. Call by \(P\) the point where the line through \(A\) meets the original line and by \(R\) the point where the other parallel line meets the original line.
2. The points \(P\) and \(R\) were already cross-aligned at step 1. Cross-align them the other way and draw along the side of the ruler through \(P\).
3. Side-align \(A\) and \(R\) and draw along the opposite side of the ruler from \(A\). You only need enough of this line to see where it meets the original line \(PR\). Call this meeting point \(X\).
4. The points \(X\) and \(R\) were already cross-aligned at step 3. Cross-align them in the other direction and draw along the side of the ruler through \(R\). Call by \(B\) the point where this line meets the line drawn through \(P\) at step 2.
5. Draw the line \(AB\). Make sure it’s long enough to meet the line through \(R\) you drew at step 1. Call the meeting point \(L\)
6. The points \(A\) and \(L\) were cross-aligned at step 5. Cross-align them in the other direction and draw along the side of the ruler through \(A\).
7. Side-align the line you just drew so that the ruler is on the opposite side from \(L\), and draw along the other side of the ruler. You only need enough of this line to see where it meets \(LR\). Call this meeting point \(M\).
8. Draw the line through \(M\) and \(A\).
Done! The line \(AB\) passes through \(A\) and is perpendicular to the original line \(PR\). The line \(AM\) passes through \(A\) and is parallel to the original line.
Video here

The point of this is that if you are going to string different constructions together to achieve a particular goal, you can choose your constructions carefully and also sometimes reuse lines from earlier parts in later parts to make the whole process a little more efficient and less full of overlapping lines everywhere.

In a long complicated construction, I would choose one or the other of these two orders for making a parallel and perpendicular line, or possibly something different entirely, depending on what existing lines and points I already had at my disposal.

This will come up in the next blog post, which is about the final two constructions that prove the two-sided ruler can do anything a ruler and compass can do.

Before I do that, though, I want to share three more constructions involving parallel lines, which I found when trying to make the constructions in the next blog post easier to do.

Draw a parallel line exactly halfway between two existing parallel lines
0. Start with two parallel lines.
1. Angle the ruler so it meets both parallel lines and draw along both sides of the ruler.
2. Side-align one of these new lines on the other side and draw along the other side of the ruler to make a third line parallel to the two drawn at step 2.
3. Two identical parallelograms have been created. Draw both diagonals of both parallelograms, creating one point in each where the diagonals meet.
4. Join these two points.
Done! The line drawn at step 4 is parallel to both original lines and halfway between them.
Video here

Proof:

The diagonals of a parallelogram bisect each other, so the two points where the diagonals of the two parallelograms meet are each halfway between the two original parallel lines.

End proof!

You don’t even need the two parallelograms to be next to each other or the same as each other and it will still work. It’s just that you can save a line by putting the parallelograms next to each other.

This construction is particularly nice if you already have a line segment meeting both parallel lines in a right angle, because then the middle parallel line must be the perpendicular bisector of the line segment.

If your line segment is longer than the ruler width, then you can very quickly draw all three of the parallel lines from the previous picture. This is the same as picking a specific distance apart for your parallel lines and also drawing one halfway between.

Draw two parallel lines a specific distance (longer than ruler width) apart and the parallel line halfway between them
0. Start with a line segment longer than the ruler width. Call the endpoints \(A\) and \(B\)
1. Cross-align the line segment and draw along the side of the ruler through \(A\).
2. Cross-align the line segment in the other direction and draw along the side of the ruler through \(B\). Call by \(X\) the point where these two lines cross.
3. Side-align \(AX\) and draw along the other side of the ruler. Call by \(P\) the point where this line meets \(BX\).
4. Side-align \(BX\) and draw along the other side of the ruler. Call by \(Y\) the point where this line meets the line drawn at step 3. Call by \(Q\) the point where this line meets \(AX\).
5. Draw \(AP\), \(BQ\) and \(XY\).
Done! The three lines \(AP\), \(BQ\) and \(XY\) are parallel, equally spaced, and perpendicular to \(AB\).
Video here

Proof:

If you imagine drawing in the other sides of the ruler at step 1 and 2, then you will create four identical rhombuses joined together into one big rhombus. The opposite-direction diagonals of these rhombuses are all at right angles to each other.

End proof!

Isn’t that lovely? Even if you don’t want the perpendicular bisector, i’s the quickest way I can think of to draw the perpendiculars at both ends of a line segment. And if you only want the perpendicular bisector, it uses exactly the same number of lines as the construction all the way in the first blog post for bisecting a line segment longer than the ruler width.

It’s worth noting that the the line joining the places where \(PY\) and \(QY\) meet \(AP\) and \(BQ\), the line \(PQ\) and the line \(AB\) are also three equally spaced parallel lines, which I think is nice. Not least because it now looks like the construction above for finding the line halfway between two parallel lines.

Of course this only works if \(AB\) is long enough to cross-align. If it’s too short, then you just have to construct the perpendiculars separately. Though if you pay attention, you can reuse quite a few of those lines…

Draw two parallel lines a specific distance (shorter than ruler width) apart and the parallel line halfway between them
0. Start with a line segment, which can be shorter than the ruler width. Call the endpoints \(A\) and \(B\)
1. Side-align the line segment and draw along that side of the ruler to extend the line segment in both directions.
2. Align \(A\) on one side of the ruler so that the opposite side meets the original line, and draw along the opposite side. Call by \(P\) the point where that line meets the original line.
3. The points \(A\) and \(P\) were cross-aligned at step 2. Cross-align them in the other direction and draw along the side of the ruler through \(A\).
4. Side-align the line you just drew so that the ruler is on the opposite side of \(A\) from \(P\) and draw along the other side of the ruler. Call by \(X\) the point where it meets the line through \(P\) drawn at step 2.
5. Draw the line through \(X\) and \(A\).
6. Align \(B\) on one side of the ruler so that the opposite side meets the original line, and draw along the opposite side. Call by \(Q\) the point where that line meets the original line.
7. The points \(B\) and \(P\) were cross-aligned at step 5. Cross-align them in the other direction and draw along the side of the ruler through \(B\). Call by \(K\) the point where this line meets \(AX\).
8. Side-align the line you just drew so that the ruler is on the opposite side of \(B\) from \(Q\) and draw along the other side of the ruler. Call by \(Y\) the point where it meets the line through \(Q\) drawn at step 5, and call by \(L\) where it meets \(AX\).
9. Draw the line through \(Y\) and \(B\). Call by \(M\) the point where it meets the line through \(A\) drawn at step 3, and call by \(N\) the point where it meets the line through \(X\) drawn at step 4.
10. Draw \(LB\) and \(YK\). Call by \(S\) the point where they meet. Also draw \(AN\) and \(MX\). Call by \(T\) the point where they meet.
11. Draw the line through \(S\) and \(T\).
Done! The three lines \(AX\), \(BY\) and \(ST\) are parallel, equally spaced, and perpendicular to \(AB\).
Video here

If you count the lines drawn here, it’s fourteen. Considering the processes I made for drawing the perpendicular bisector of a line shorter than a ruler width both took eleven lines, I think this is pretty good! It’s only got three extra lines and two of those were part of the goal!

That concludes my discussion of parallel (and perpendicular) lines. The next blog post is about circles.