Two-sided ruler constructions 4: Copying and cutting

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

This blog post is about copying line segments and cutting them into parts, something that until now we’ve only been able to do with line segments if they’re long enough. To be able to do these constructions, we’re going to need some help from similar and congruent triangles, which were the standard way to prove almost everything in classical ancient Greek geometry.

The first construction is for copying a line segment along the line it is part of.

Copy a line segment next to itself
0. Start with a line segment of any length that is part of a longer line. Call the endpoints of the line segment \(A\) and \(B\)
1. 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.
2. Draw a line through \(A\) so that it meets both parallel lines on either side. Call these points \(P\) and \(Q\).
3. Draw the line \(PB\) and find where it meets the parallel line through \(Q\). Call this point \(X\).
4. Draw the line \(QB\) and find where it meets the parallel line through \(P\). Call this point \(Y\).
5. Draw the line \(XY\) and find where it meets the extended line segment. Call this point \(C\).
Done! \(BC\) is the same length as \(AB\).
Video here

This procedure is in Wernick, and I laughed out loud when I saw it because it is so clever. Here’s Wenrick’s proof in my own words.

Proof:

Triangles \(\triangle APB\) and \(\triangle QPX\) are similar because they share the angle at \(P\) and the other edges are parallel. Therefore their matching edges are in the same proportion.

Since the three parallel lines are the same distance apart, that means \(PA\) is half the length of \(PQ\). Hence \(AB\) is half the length of \(QX\).

By similar reasoning, triangles \(\triangle BYC\) and \(\triangle QYX\) are similar and \(BC\) is half as long as \(QX\).

Therefore \(BC\) is the same length as \(AB\), since they’re both half of \(QX\).

End proof!

Isn’t that clever? I love so much how you make a length twice as long and then make one half as long as that. Also the diagram is so neat and spacious. And you can just repeat it as many times as you want to multiply the line segment by any natural number.

It’s worth noting that the process will work the same even if the three parallel lines aren’t equal distances apart. The proportion between the segments on the middle line and the segments on the outside lines will be the same as the proportion between the distances between the lines, so the two line segments will still work out the same as each other. So, if you happen to already have parallel lines on either side of the original line segment, you don’t have to draw new ones a ruler width away.

It even works when the two extra parallel lines are on the same side of the original line, rather than on opposite sides, though the picture isn’t nearly so symmetrical.

It’s also worth noting that you can actually start the new equal-length segment at any point on the extended line \(AB\), by using that point in the place of \(B\) at step 4. That’s why Wernick called this one copy a line segment instead of double.

Copy a line segment along its line to an arbitrary point
0. Start with a line segment of any length that is part of a longer line, and another point. Call the endpoints of the line segment \(A\) and \(B\), and the other point \(C\).
1. 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.
2. Draw a line through \(A\) so that it meets both parallel lines on either side. Call these points \(P\) and \(Q\).
3. Draw the line \(PB\) and find where it meets the parallel line through \(Q\). Call this point \(X\).
4. Draw the line \(QC\) and find where it meets the parallel line through \(P\). Call this point \(Y\).
5. Draw the line \(XY\) and find where it meets the extended line segment. Call this point \(D\).
Done! \(CD\) is the same length as \(AB\).
Video here

There is another process for doubling a line segment mentioned in Wernick and Birrel. I’m describing it a little differently here.

Double a line segment
0. Start with a line segment of any length.
1. Side-align the line segment and draw along that side of the ruler to extend the segment in the direction you want to double it. Call the enpoint of the line segment on the extended side \(B\) and the other endpoint \(A\).
2. Side-align the extended line segment and draw along the other side of the ruler to make a parallel line.
3. Draw a line through \(A\) that meets the parallel line you just drew. Call the point where it meets that line \(P\).
4. Align the point \(P\) on the side of the ruler so that the opposite side of the ruler is on the same side as \(B\). Draw along the opposite side of the ruler to make a parallel line. You need enough of the line to be able to side-align it later and to find where it meets the line drawn at step 2. Call this meeting point \(Q\). Note it will work best if you make sure \(PQ\) is longer than \(AB\).
5. Side-align the line drawn in step 4 on the other side, and draw along the opposite side of the ruler to make a parallel line. You only need enough of this line to find where it meets the line drawn at step 2. Call this meeting point \(R\).
6. Draw the line \(QB\) and find where it meets the line \(PA\). Call this meeting point \(X\).
7. Draw the line \(XR\) and find where it meets the extended original line segment. Call this meeting point \(C\).
Done! \(BC\) is the same length as \(AB\).
Video here

Proof:

Triangles \(\triangle AXC\) and \(\triangle PXR\) are similar because they share the angle at \(X\) and the opposite edges are parallel. Therefore the proportions \(AX:PX\) and \(CX:RX\) are the same.

Triangles \(\triangle AXB\) and \(\triangle PXQ\) are also similar for the same reason as before. Therefore the proportions \(AX:PX\) and \(AB:PQ\) are the same.

Triangles \(\triangle BXC\) and \(\triangle QXR\) are also similar for the same reason as before. Therefore the proportions \(CX:RX\) and \(BC:QR\) are the same.

Therefore the proportions \(AB:PQ\) and \(BC:QR\) must be the same. But the lengths \(PQ\) and \(QR\) are the same which means that \(AB\) and \(BC\) must be the same length too, to keep the same proportions.

End proof!

I love how we solved the problem of doubling a segment that’s too short by making a different segment that’s longer than the ruler width, doubling that, and relating the two together. Indeed, in Wernick and Birrell, they actually describe the process by saying to just create three equally spaced points on that first parallel line you drew using the process for lines longer than the ruler width. I wanted to include it all in one process.

If you want it to turn out the most nicely, you do have to decide on how widely spaced those three points are based on how long the original segment is. You’ll probably only use this process if the segment is shorter than the ruler width, in which case it won’t really matter that much, but you never know, so I’m just telling you just in case.

It’s worth noting again that this will work fine if your parallel line isn’t one ruler width away, so if you happen to already have a parallel line, you can use it just fine. It will also work if you just happen to already have three equally spaced points on that parallel line. Note, though, that if the spacing is shorter than the original segment, then the point \(X\) will be on the other side of the parallel line.

Finally, just like last time, you can modify this process to make as many copies of \(AB\) as you like by first making more copies of \(PQ\).

This process can also be modified to divide a segment into equal parts, instead of multiplying it.

Bisect a line segment
0. Start with a line segment of any length. Call the endpoints \(A\) and \(C\).
1. Side-align the line segment and draw along the opposite side to make a parallel line.
3. Draw a line through \(A\) that meets the parallel line you just drew. Call the meeting point \(P\).
4. Align the point \(P\) on the side of the ruler so that the opposite side of the ruler is on the same side as \(C\). Draw along the opposite side of the ruler to make a parallel line. You need enough of the line to be able to side-align it later and to find where it meets the line drawn at step 2. Call this meeting point \(Q\). Note it will work best if you make sure \(PQ\) is longer than \(AB\).
5. Side-align the line drawn in step 4 on the other side, and draw along the opposite side of the ruler to make a parallel line. You only need enough of this line to find where it meets the line drawn at step 2. Call this meeting point \(R\).
6. Draw the line \(RC\) and find where it meets the line \(PA\). Call this meeting point \(X\).
7. Draw the line \(XQ\) and find where it meets the extended original line segment. Call this meeting point \(B\).
Done! \(AB\) is the same length as \(BC\).
Video here

The proof is identical to the proof of the previous construction, because except for the lengths of the segments involved, the diagram is identical to the diagram in the previous construction, even though we drew the lines in a slightly different order.

Again if you happen to have an existing line parallel to the segment, you can use that and it will still work. And if you happen to have equally spaced points on that parallel line, you can use them instead of making your own (though the point \(X\) may be on the other side of the parallel line if the points are too close together.) And just like before you can modify it to make as many copies of \(PQ\) as you want before joining \(C\) to the outermost point, and so divide the segment into as many pieces as you desire.

If you count the number of lines drawn to do this construction, including making the parallel line and the equally spaced points, you’ll get a total of six lines. This is one more than the number of lines it took to perpendiculalry bisect a line segment longer than the ruler width by drawing a rhombus around it. However, if you already have a parallel line or a parallel line and some points marked on it, this process will be much quicker because you won’t have to draw those lines yourself.

I find this “line saving” calculation fascinating. You can make arguments about a procedure being equally or less efficient than another, but in the right context, it might be more efficient because of being able to reuse already-drawn lines. It feels so satisfying to me to find these moments in the more complicated constructions.

There’s only one more construction I want to do in this post, which is to copy a length from one arm of an angle to the other. Both Wernick and Birrell have the same construction for this, which involves drawing a perpendicular line (a construction I’ve put later). But Chris Tisdell has a different construction, which is more direct and I like it very much. That’s the one I present here, except I don’t draw all the lines he draws.

Copy a length from one arm of an angle to the other
0. Start with two lines forming an angle, and a line segment along one of the arms starting at the vertex. Call the vertex of the angle \(P\), and the other end of the line segment \(A\).
1. Draw a rhombus with the angle as one of its angles and bisect that angle, as described earlier. Call by \(Q\) the vertex of the rhombus on same arm as \(A\), call by \(R\) the vertex of the rhombus between the two angle arms, and call by \(S\) the vertex of the rhombus on the other arm.
2. Draw the line \(AS\) and find where it meets the angle bisector \(PR\). Call this point \(X\).
3. Draw the line \(QX\) and find where it meets the second arm of the angle \(PS\). Call this point \(B\).
Done! \(PB\) is the same length as \(PA\).
Video here

Proof:

Consider triangles \(\triangle PSX\) and \(\triangle PQX\).

Side \(PS\) is the same length as side \(PQ\), since they’re two sides of a rhombus.

Angle \(\angle SPX\) is the same as \(\angle QPX\) since we set up to bisect the original angle.

Side \(PX\) is shared.

Therefore \(\triangle PSX\) and \(\triangle PQX\) are congruent.

Therefore \(\angle PXS\) is the same as \(\angle PXQ\).

Also note that \(\angle SXB\) is the same as \(\angle QXA\) since they’re vertically opposite.

Adding these angles together, \(\angle PXB\) is the same as \(\angle PXA\).

Now consider triangles \(\triangle PXB\) and \(\triangle PXA\).

We just showed \(\angle PXB\) is the same as \(\angle PXA\).

We already said \(\angle BPX\) is the same as \(\angle APX\) since we set up to bisect the original angle.

Side \(PX\) is still shared.

Therefore \(\triangle PXB\) and \(\triangle PXA\) are congruent.

Hence length \(PB\) is the same as length \(PA\).

End proof!

I had a lot of fun making this proof, since I hadn’t done triangle congruence proofs for a while. I’m sure there are other ways to do it, but this is mine.

I was a bit worried how it might look when the length of \(PA\) was shorter than the side length of the rhombus. Here’s the diagram in that case, with all the point labels done as described in the construction.

If you follow the proof with this diagram, all the arguments work exactly the same, except for one moment. In the previous proof, you know \(\angle PXB\) is the same as \(\angle PXA\) because they were made by joining other angles together. In this case, you know they’re the same because you subtract angles from each other. But it still works.

There’s something very interesting about these diagrams too. The triangle \(\triangle APB\) is isosceles, which means the bisector of the angle at \(P\) will meet the opposite side \(AB\) in a right angle. So we haven’t just copied the length \(PB\) to the other side of the angle, we’ve also dropped a perpendicular from \(A\) to the line \(PR\). This is going to be very important in the next blog post, which is about perpendicular lines.