Two-sided ruler constructions 5: Perpendicular lines

This is the fifth 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 (you are here)
6. Parallel lines
7. Circles without circles
8. Equilateral triangle and regular pentagon

This blog post is all about drawing perpendicular lines exactly in the place you want them. It feels surprising that you could do these with just the two-sided ruler, but it is very much possible.

I’ve already made perpendicular lines a few times. In the post about rhombuses, I had two constructions of a right angle just anywhere, and also the perpendicular bisector of a line segment longer than a ruler width. And in the post about copying and cutting, I noticed that the construction for copying a length from one arm of an angle to another had some perpendicular lines in it. In this post I’ll be taking inspiration from all of these to put perpendiculars in very specific places.

The first construction is to draw a line perpendicular to an existing line through a point on that line. This is my construction and actually you’ve seen it before when I made a right angle just anywhere, it’s just that this time I have one of the lines already made.

Draw a line perpendicular to an existing line through a point on that line
0. Start with a line and a point on that line. Call the point \(B\).
1. 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 \(A\) the point where that line meets the original line.
2. The points \(A\) and \(B\) were cross-aligned at step 1. Cross-align them in the other direction and draw along the side of the ruler through \(B\).
3. Side-align the line you just drew so that the ruler is on the opposite side of the line from \(A\) and draw along the other side of the ruler. Call by \(C\) the point where this line meets the original line and call by \(X\) the point where it meets the line through \(A\) drawn at step 1.
4. Draw the line through \(X\) and \(B\).
Done! The angle \(\angle ABX\) is a right angle.
Video here

The diagram here is the exact same one for the way to make a right angle using a triangle, so the proof that the angle is a right angle is exactly the same too.

I deeply love this construction. It takes exactly four lines including the perpendicular line you were hoping to draw. You don’t even need to make that third line long enough to meet the original line at point \(C\), just make a little mark where the ruler meets the line through \(A\). I only needed to talk about point \(C\) to make my proof easier to describe.

Wernick and Birrell don’t do this construction. Instead they make two points \(A\) and \(C\) on either side of \(B\) so that \(B\) is their midpoint, and then draw a rhombus with diagonal \(AC\) and draw in the other diagonal. But I noticed that the very lines you need to draw to create those points on either side of \(B\) already show you where the perpendicular should be. You don’t need to draw any more!

If you happen to already have the middle point from three equally spaced points, and the outside two are further than a ruler width apart, then I concede you can very quickly find the perpendicular by drawing a rhombus. But you only need half the rhombus.

Draw a line perpendicular to an existing line through the middle point of three equally-spaced points on the line
0. Start with a line and three equally spaced points on the line so that the outside two are further than a ruler width apart. Call the three points in order \(A\), \(B\) and \(C\).
1. Cross-align \(A\) and \(C\) and draw along the side of the ruler through \(A\).
2. Cross-align \(A\) and \(C\) in the other direction and draw along the side of the ruler through \(C\). Call by \(X\) the point where the two lines drawn at step 1 and 2 meet.
3. Draw the line through \(X\) and \(B\).
Done! The angle \(\angle ABX\) is a right angle.
Video here

Proof:

The sides \(AX\) and \(CX\) of triangle \(\triangle AXC\) were created by cross-aligning \(AC\) so they both meet \(AC\) in the same angle by the cross-align arcsine lemma. So, \(\triangle AXC\) is an isosceles triangle and the line joining \(X\) and the midpoint of \(AC\) must meet \(AC\) in a right angle.

End proof!

Again, if I had drawn in both sides of the ruler when I cross-aligned, I would have drawn a rhombus and could have used the fact that its diagonals bisect each other at right angles. But I don’t need to draw the whole rhombus! I can save two lines and do this in just three lines, which makes me happier than it probably has a right to.

At this point, I think I should address the fact that we have a method for finding the perpendicular bisector of a line segment longer than the ruler width, but not one for shorter line segments.

Since we do have a way to bisect short segments and a way to draw a perpendicular through a specific point, it can definitely be done. It takes six lines to bisect a segment and four lines to draw a perpendicular to a line through a point on the line, so that’s ten lines in total. Except you’ll have to extend the line segment to draw that perpendicular, so that’s eleven lines total. Try as I might, I just cannot find a quicker way than that, so I’m just going to leave it at that.

Draw the perpendicular bisector to a line segment
0. Start with a line segment. Call the endpoints \(A\) and \(C\).
1. Bisect the line segment by the process shown before. That is: – Draw a line parallel to the segment – Mark three equally spaced points on the line and connect the outside ones to \(A\) and \(C\) – Connect the middle point to the intersection point just created and find where that line meets the segment. Call the midpoint \(B\).
2. Side-align the segment \(AB\) and draw along that side of the ruler to extend the segment. You only need to extend it in one direction (even though in the video I extended it in both.)
3. Draw a perpendicular line to the segment through \(B\) the midpoint using the process shown before. That is – Cross-align \(B\) and some other point \(P\) on the extended segment and draw the along the side of the ruler through \(P\). – Cross-align in the other direction and draw along the side of the ruler through \(B\). – Side-align that line and draw along the other side of the ruler making a line meeting the line just drawn through \(P\) at \(X\). – Draw \(XB\).
Done! The line \(XB\) is the perpendicular bisector of \(AC\).
Video here

Actually I think I will share one of the other ways I found to draw the perpendicular bisector of a line segment using eleven lines that I find rather charming.

Draw the perpendicular bisector to a line segment (by copying a length first)
0. Start with a line segment of any length. Call its endpoints \(A\) and \(B\)
1. Side-align the segment \(AB\) and draw along that side of the ruler to extend the segment in both directions.
2. Align \(A\) on one side of the ruler in such a way that the other side of the ruler meets the extended line segment beyond \(B\), and draw along both sides. Call by \(C\) the point where the line not through \(A\) meets the extended line segment.
3. Side-align the line segment and draw along the opposite side of the ruler to make a parallel line. Repeat on the other side of the line segment. Call by \(P\) and \(Q\) the points where the line through \(A\) drawn at step 2 meets these parallel lines. It works best if \(P\) is the point on the opposite side of \(A\) from \(B\).
4. Draw the line through \(P\) and \(C\) so that it meets the parallel line through \(Q\). Call this meeting point \(X\).
5. Draw the line through \(X\) and \(B\) so that it meets the parallel line through \(P\). Call this meeting point \(Y\).
6. Draw the line through \(Q\) and \(Y\) so that it meets the extended line segment \(AB\). Call this meeting point \(Z\).
7. Cross-align the points \(Z\) and \(B\) so that the sides of the ruler are not parallel to \(PQ\) and draw along both sides of the ruler. Call by \(L\) the point where the line just drawn through \(B\) meets \(PQ\), and by \(M\) the point where the line just drawn through \(Z\) meets the line drawn through \(C\) at step 2.
8. Draw the line through \(L\) and \(M\).
Done! The line \(LM\) is the perpendicular bisector of \(AB\).
Video here

Proof:

Steps 2 to 6 are the construction for copying line segment \(AC\) to \(ZB\). Removing \(AB\) from both line segments must produce the same lengths, so \(ZA\) is the same length as \(CB\).

Since \(ZB\) and \(AC\) are the same length, cross-aligning \(ZB\) will produce lines that meet the extended line segment \(ZC\) in the same angles as when cross-aligning \(AC\).

This means \(ZM\) and \(CM\) meet \(ZC\) in the same angle and so \(ZMC\) is an isosceles triangle. This means the perpendicular bisector of \(ZC\) will pass through \(M\).

Also \(AL\) and \(BL\) meet \(AB\) in the same angle and so \(ALB\) is an isosceles triangle. This means the perpendicular bisector of \(AB\) will pass through \(L\).

However, \(AB\) and \(ZC\) share a centre, since \(ZA\) is the same as \(CB\), and so they have the same perpendicular bisector.

Both \(L\) and \(M\) are on that perpendicular bisector, so \(LM\) is the perpendicular bisector.

End proof!

One reason I like this construction is that the process for longer line segments also begins with cross-aligning some points on the line segment (the endpoints), so it’s as if we’ve tried to do that process and then when it didn’t work we pivoted to something new. And yet by the end there was a rhombus anyway and the perpendicular bisector was its diagonal after all.

Next, we want to draw a line perpendicular to an existing line through a point not on the line. In both Wernick and Birrell, they do this by first drawing a perpendicular through a point on the line, and then drawing a line parallel to that one through the point in question. (Drawing parallel lines is in the next blog post, but Wenick and Birrell do parallel lines before perpendicular lines.) When reading that, I was sure there had to be a quicker way. And there is.

Draw a line 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 \(A\).
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 line. 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 in the other direction and draw along both sides of the ruler. This creates a rhombus with \(PR\) as a diagonal. Call by \(Q\) the vertex of the rhombus on \(PA\) and by \(S\) the final vertex of the rhombus.
3. Draw the line \(AS\) and find where it meets the rhombus diagonal \(PR\). Call this point \(X\).
4. Draw the line \(QX\) and find where it meets the line \(PS\). Call this point \(B\).
5. Draw the line \(AB\).
Done! The line \(AB\) passes through \(A\) and meets the original line \(PR\) in a right angle.
Video here

Proof:

The diagram here is the exact same diagram as when we copied a length from one arm of an angle to the other in the previous blog post. Therefore \(PA\) is the same length as \(PB\) and \(\triangle APB\) is an isosceles triangle.

Hence, the angle bisector of the angle at \(P\) must meet the opposite side \(AB\) in a right angle.

End proof!

Note there’s a bit of ambiguity in my description as to which side of \(A\) you put the ruler and therefore which side of the line \(AP\) the point \(R\) is. The reason I didn’t address it is that the construction works perfectly well either way. This is what it looks like if the ruler is on the other side of \(A\).

The reason I didn’t present it this way is that if you’re not careful, the line \(AS\) might take a very long distance to actually meet the original line, so it’s safest to do it the other way.

As I noted in the proof, this is pretty much exactly the method of of copying a length from one side of an angle to the other based on Tisdell’s method. However, I actually created this method before seeing Tisdell’s method of copying a line segment and then noticed how similar they were afterwards. When I made this, I was trying to find a way to do what Wernick and Birrell told me to do and make a perpendicular line anywhere and then a line parallel to that line through \(A\), while also trying to reuse as many lines as possible rather than draw new ones at every step, and also avoid drawing lines that weren’t necessary. Though I didn’t draw it, the first perpendicular is \(QS\) when that rhombus was made. And you can see that \(AB\) is indeed parallel to \(QS\).

Seeing it now, I wonder if you could make a line parallel to an existing line through a specific point by putting this diagram in the right location. You can, and it makes a neat connection between perpendiculars and parallels that helped me decide to tell the story in the order I have, rather than do parallels first like Wernick and Birrell did.

Before I do that, I have to show you another way to find a line perpendicular to another through a point not on the line. I came up with it while writing a part of the next blog post and I’ve come back here to put it in because I like it so much. It only draws six lines total, as opposed to seven.

Draw a line perpendicular to an existing line through a point not on that line (without drawing a rhombus)
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. 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 \(R\) (it doesn’t matter which side), 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 \(AB\).
Done! The line \(AB\) passes through \(A\) and meets the original line \(PR\) in a right angle.
Video here

Proof:

Consider the triangles \(\triangle APR\) and \(\triangle BPR\).

The lines \(AR\) and \(BR\) meet \(PR\) in the same angle since they were produced by cross-aligning \(PR\), so \(\angle ARP = \angle BRP\).

The lines \(AP\) and \(BP\) meet \(PR\) in the same angle since they were produced (or at least imagined) while cross-aligning \(XR\), so \(\angle APR = \angle BPR\).

Finally the side \(PR\) is shared. Therefore \(\triangle APR\) and \(\triangle BPR\) are congruent.

Hence \(AP = BP\) and the triangle \(\triangle APB\) is isosceles. Since the angles between \(PR\) and both of \(AP\) and \(BP\) are the same, \(PR\) is the angle bisector of the angle at \(P\) in isosceles triangle \(\triangle APB\), therefore it must meet the base \(AB\) in a right angle.

End proof!

It’s worth showing what this looks like when you place the ruler on the other side of the point \(A\).

Whichever one you choose, they both tend to require a lot of space if the point \(A\) is close to the original line, so fair warning.

I was inspired to create this one by thinking about the isosceles triangles that made some of the other constructions work, and wondering if perhaps I could get the perpendicular I wanted by constructing two different isosceles triangles, or if you like, constructing a kite.

I do love all the cross-aligning that’s happening here. The saving of an extra line might not feel much of a saving compared to the care with which you have to enact all that precise ruler work. Also the extra space required when \(A\) is close to the line might prohibit using it too. It’s up to you to decide what works best for you.

And before anyone says it, I have indeed drawn three of the four sides of a rhombus with diagonal \(PR\), but I didn’t draw the fourth one so I still think it counts as not drawing a rhombus.

That completes the ways to make lines at right angles to each other. The next blog post is about how to make lines parallel to each other.